Solutions Solved Exercise

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   Solved exercise of chapter 9 solutions form chemistry book 1 for FSC pre engineering and pre medical of Board of intermediate and secondary education.                        




                         TEXT BOOK EXERCISE
Q1.      Choose the correct answer for the given ones.
(i)                 Morality of pure water is
(a)        1.         (b)        18.       (c)        55.5     (d)       6.
Hint:    Morality of pure water
Consider 1 dm3 (-1000cm3 ) of water. Convert this volume into mass by using density .                
Mass =volume x density
            Mass of H2O = 1000 cm3 x 1 g cm3 =1000g
Molar mass of H2O     =18 g mol-1
No. of moles of H2O   ==55.6 mol                     
Morality of H2O          =
                                    ==55.6 mol dm-3 =55.6 M
(ii)        18 g glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to
             (a)        1          (b)        18        (c)        55.5     (d)       6
(iii)       A solution of glucose is 10%. The volume in which 1 g-mole of it is dissolved will be
            (a)        1 dm3   (b)        1.8dm3 (c)        200 cm3               (d)       900cm3
(iv)       An aqueous solution of ethanol in water may have vapour pressure
            (a)        equal to that of water              (b)        equal to that of ethanol
            (c)        more than that of water          (d)       less than that of water
(v)        An azeotropic mixture of two liquids boils at a lower temperature than either of them when
            (a)        it is saturated
            (b)        it shows positive deviation from Raoult’s law
            (c)        it shows negative deviation from Raoult’s law
            (d)       it is met stable
(vi)       In azeotropic mixture showing positive deviation from Raoult’s law , the volume of the mixture is
(a)                slightly more than the total volume of the components
(b)               slightly less than the total volume of the components
(c)                equal to the total volume of the components
(d)               none of these
(vii)           Which of the following solution has the higher boiling point?
(a)                5.85% solution of sodium chloride
(b)               18.0% solution of glucose
(c)                6.0% solution of urea
(d)               All have the same boiling point.
(e)                Two solution of NaC1and KC1 are prepared separately by dissolving same amount of solute in water. Which of the following statements is true for these solutions?
(f)                KC1 solution will have higher boiling point than NaC1 solution
(g)               Both the solution have different boiling points.
(h)               KC1 and NaC1 solutions possess same vapour pressure
(i)                 KC1 solution possesses is the ratio of the elevation in boiling point to
(j)                 The molal boiling point constant is the ratio of the elevation in boiling point to
(a)        molaritly                                  (b)        molality
(c)        mole fraction of solvent          (d)       mole fraction of solute
(x)        Colligative properties are the properties of
(a)        Dilute solutions which behave as nearly ideal non-ideal solutions
(b)        Concentrated solutions which behave as nearly non-ideal solutions
            (c)        Both (a) and (b)
            (d)       Neither (a) and (b)
Ans.    (i)         c          (ii)        c          (iii)       b          (iv)       c          (v)        b
            (vi)       a          (vii)      a          (viii)     c          (ix)       b          (x)        a

Q2.      Fill in the blanks with suitable words or numbers.

(i)                 Number of molecules of sugar in 1dm3 of 1 M sugar solution is _________.
(ii)               100 g of a 10% aqueous solution of NaOH contains 10 g NaOH in _________ g of water.
(iii)             When an azeotropic mixture is distilled, its __________remains constant.
(iv)             The molal freezing point of an azeotropic solution of  two liquids is lower than ether of them because the solution shows________from Raoluls’s law.
(v)               Among equimolal aqueous solution of NaC1, BaC12, and FeC13, the maximum depression in freezing point is shown by _______solution.
(vi)             A solution of ethanol in water shows____deviations and gives azeotropic solutions with _______b.p. than other components.
(vii)           Colligative properties are used to calculate _______of a compound.
(viii)         The boiling point of an azeotropic solution of two liquids is lower than either of them because the solutions show ________from Raoult’s law.
(ix)             The hydration energy of Br- ion is _________than that of F- ion.
(x)                 The aqueous solution of NH4C1 is _________while that of Na2SO4 is ­­­­_______.

Ans.    (i)6.02x 1023                (ii)90                (iii)composition                       (iv)cryoscopic (v)Positive deviation  
(vi)FeC13                     (vii)positive ; lower    
(viii) molar mass          (ix) lesser                     (x) acidic ;neutral

Q3.      Indicate True or False from the given statements.

(i)                 At a definite temperature the amount of a solute in a given saturated solution is fixed.
(ii)               Polar solvents readily dissolve non-polar covalent compounds.
(iii)              The solubility of a substance decreases with increase in temperature if the heat of a solution is negative.
(iv)             The rate of evaporation of a liquid is inversely proportional to the intermolecular forces of attraction.
(v)               The molecular mass of an electrolyte determined by lowering of vapour pressure is less than the theoretical molecular mass.
(vi)             Boiling point elevation is directly proportional to the molality of the solution and inversely proportional to boiling point of solvent.
(vii)           All solutions containing 1 g of non-volatile non-electrolyte solutes in some solvent will have the same freezing point.
(viii)         The freezing point of a 0.05 molal solution of a non-volatile non-electrolyte solute in water is – 0.93oC.
(ix)             Hydration and hydrolysis are different process for Na2SO4.
(x)               The hydration energy of an ion only depends upon its charge.
Ans.    (i) True            (ii)False           (iii)True           (iv)True           (v)True
            (vi) True          (vii) False        (viii) False       (ix)True           (x) False
Q4.      Define and explain the following with examples:
(a)        A homogeneous phase            (b)        A concentrated solution
(c)        A solution of solid in a solid   (d)       A consulate temperature
(e)        A non-ideal solution               (f)        Zeotropic solutions
(g)        Heat of hydration                   (h)        Water of crystallization
(i)         Azeotropic solution                 (j)         Conjugate solution
Q5.      (a)        What are the concentration units of solutions? Compare molar and molal solution.
Ans.    Compare molar and molal solution.
           
Molar Solution
Molar Solution
1.   “A solution which contains one mole of solute per dm3 of solution is called a molar solution.”
2.  Its morality decreases with the rise in temperature.
3.  Its unit is mole dm3 of solution.
1.   “A solution which contains one mole of solute per kilogram of solvent is called a molal solution.”
2. Its molality is independent of temperature variation of solution.
3.   Its unit is mol kg-1 of solvent.
           
(b)       One has one molal solution of NaC1 and one molal solution of glucose.
            (i)         Which solution has greater number of particles of solute?
            (ii)        Which  solution has greater amount of the solvent.?
(iii)       How do we convert these concentrations into weight by weight percentage?
(b)       (i)         One molal solution of NaC1 has greater number of particles of solute.
            (ii)        The amount of solvent in both solution is equal (1 kg)
            (iii)                   Mass of NaC1 = 58.5 g
                        Mass of solvent           =1000g
                        Total mass of NaC1 solution =1000 + 58.5 =1058.5 g
                                    Percentage of NaC1 =x100
                                                =x100=5.53 % Answer
                        Mass of glucose          =180g
                        Mass of solvent           =1000g
                        Total mass of solution =1000+180=1180 g
                                    % of glucose   =x100
                                                            =x100=15.25 % Answer
Q.6      Explain the following with reasons
(i)         The concentrations in terms of molality is independent of the temperature but morality depends upon temperature.
Ans.    Molality is based on the mass of solvent. The mass of solvent does not vary not vary with temperature, so molality is independent of temperature. Morality is based on the volume of solution. Since the volume of solution varies with temperature, so morality depends upon temperature. The morality of the solution decreases with the increase in temperature of the solution.
(ii)        The sum of the mole fractions of all the components is always equal to unity for any solution.
Ans.    Suppose there be two components A and B making a solution. The numbers of moles are nA and nB respectively, then. If the mole fractions of A and B are denoted by XA and XB respectively, then
                                    XA = and XB =
            The sum of mole fractions of the components of a solution will be
                                    XA + XB  =+
                                    XA + XB  =                        
                                    XA + XB  =1
Hence, the sum of the mole fractions of all the components of any solution is always equal to unity.
(iii)       100 g of 98% H2SO4 has a volume of 54.34 cm3 of H2SO4 since its density is 1.84 g cm-3.
Ans.                Mass of 98% H2SO4 =100g : density of 98% =1.84 g cm-3
                        Vol. of 98% H2SO4=?
                        Formula Used:                       volume ===54.34 cm-3
            Hence, 100 g of 98 % H2SO4 has a volume of 54.34 cm-3 of H2SO4 because its density is 1.84 gcm-3 . 
(iv)       Relative lowering of vapour pressure is independent of the temperature.
Ans.    Since, the   relative lowering of vapour pressure is equal to the mole fraction of solute, so it is independent of temperature.
            Hint:               =X2          or         =
            (v)        Colligative properties are obeyed when the solute is non-electrolyte and also when the solutions are dilute.
Ans.    Colligative properties are obeyed when the solute is non-electrolyte
            Colligative properties depend only upon the number of solute particles and not on their chemical nature. An electrolyte solute differs from a non-electrolyte principally in the number of particles produced upon dissolution. In case the solute is non-electrolyte , one mole of solute produces one mole of dissolved particles (molecules). In case the solute is electrolyte, it may split into a number of ions each of which acts as a particle and thus will affect the colligative properties of solution than non-electrolytes. For example, one mole of glucose produces one mole of dissolved particles (molecules) while one mole of NaC1 produces two moles of dissolved particles (one mole each of Na+ and C1- ions). Thus, mole for mole, the NaC1 exerts twice the colligative effect than glucose if the solution is ideal.
            Colligative properties are obeyed when the solutions are dilute.
             Colligative properties are obeyed when the solutions are dilute. A dilute solution behaves almost as an ideal solution, i.e., the solute-solute interactions are negligible. Concentrated solution is mostly non-ideal.
            (vi)       The total volume of the solution by mixing 100 cm3 of water with 100 cm3 of alcohol may not be equal to 200 cm3. Justify it.
Ans.    Because the intermolecular forces of attraction between alcohol and water molecules are not the same as the intermolecular attractive forces between alcohol molecules or between water molecules. Hence, the total volume of the solution by mixing 100 cm3 of alcohol with 100 cm3 of water will not be equal to 200 cm3. The total volume of solution will be greater than 200 cm3 because the forces of   attraction between alcohol and water molecules are weaker than those between alcohol molecules or between water molecules.
            (vii)      One molal solution of urea in water is dilute as compared to one molar solution of urea, but the number of particles of the solute is the same. Justify it.
Ans.    One molal solution of urea in water is dilute as compared to one molar solution of urea. This is because a molal solution contains one mole of urea in 1000 g of water whereas one molar solution contains one mole of urea in 1000 cm3 of water.
            At room temperature, density of water is slightly less than one. Therefore, the volume corresponding to 1000 g of water be greater than 1000cm3. So, the volume of solvent water containing one mole of solute is more in case of molal solution than molar solution. hence, one molal solution of urea in water is dilute as compared  to one molar solution of urea in water. Since both the solutions contain 1 mole of urea as solute, therefore , the number of particles of solute is the same in both the solution.
            (viii)     Non-ideal solution do not obey the Raoult’s law.
Ans.    They show deviations from Raoult’s law due to differences in their molecular structures, i.e. , size , shape and intermolecular forces. Formation of such solutions is accompanied by changes in volume and enthalpy. The vapour pressure deviation may be positive or negative in such solutions.
Q7.      What are non-ideal solutions? Discuss their types and give three example of each.
Q8.      (a)        Explain fractional distillation. Justify the two curves when composition is plotted against boiling point of solutions.
            (b)        The solution showing positive and negative deviations cannot be fractionally distilled at their specific compositions. Explain it.
Q9.      (a)        What are azeotropic mixtures? Explain them with the help of graphs.
            (b)        Explain the effect of temperature on phenol-water system.
Q10.    (a)        What are collligative properties? Why are they called so?
Ans.    Because the colligative properties of solution depend only upon the number of solute and solvent particles present in the solution and not upon the chemical nature of the solute molecules. For this reason these properties are called colligative properties.
            (b)        What is the physical significance of KB and kf values of solvent?
Ans.    Because the change in boiling point and the freezing point of a solvent is a colligative property that depends only on the ratio of the number of particles of solute an dissolvent in the solution, so these constants are used to determine the molecular mass of an unknown solute.
Q11.    How to    explain that the lowering of vapour pressure is a colligative property?
            How do we measure the molar mass of a non-volatile, non-electrolyte solute in a volatile solvent?
Q12.    How do you justify that?

            (a)        Boiling points of the solvents increase due to the presence of solutes.
Ans.    “The temperature of a pure liquid at which its vapour pressure becomes equal to an external atmospheric pressure on its surface is called boiling point of the liquid.”
            The addition of a non-volatile solute to a solvent always reduces its vapour pressure below that of the pure solvent. This is because solute particles occupy some of the surface area of the solution decreases the rate of evaporation and thus reduces its pressure. So, a higher temperature is needed to increase the vapour pressure to the point where the solution boils. Hence, the boiling points of the solvents increase due to the presence of solutes.

            (b)        Freezing points are depressed due to the presence of solutes.
Ans.    “The temperature f a pure liquid at which its solid and liquid are froms coexist in equilibrium is called freezing point of the liquid.”
            When a non-volatile solute is added to a solvent, it reduces its vapour pressure. If a solution is cooled sufficiently, the temperature at which crystals of pure solvent appear is the freezing point of the solution. At this temperature the solid solvent and solution are in equilibrium, so they must have the same vapour pressure. But , at a definite temperature, a solution containing a non-volatile solute has a lower vapour pressure than the pure solvent. Therefore, solid solvent must be in equilibrium with a solution at a lower temperature than the temperature that it would be in equilibrium with pure solvent. Hence, freezing points are depressed due to the presence of solutes.

            (c)        The boiling point of one molal urea solution is 100 .52o C but the boiling of two molal urea solution is less than 101.04oC. 
Ans.    The boiling point of 1 molal urea solution is 100.52oC at 1 atm pressure rather than 100oC. Thus the elevation of boiling point is (100.52 – 100.00) =0.52oC. The boiling point of 2 molal urea solution is 101.04 oC , so the elevation of boiling points is (101.04 – 100.00)= 1.04oC. Since the boiling point elevation depends upon the number of particles of solute, therefore, e molal urea solution which contains 2 x 6.02 x 1023 molecules has double the boiling point elevation than 1 molal urea solution which contains 6.02 x1023 molecules.

            (d)        Beckmann’s thermometer is use to note the depression in freezing point.
Ans.    Beckmann’s thermometer is used to read temperatures up to, 0.01oC over a range of about 5oC. Since, freezing point depressions are small , no more than a degree or two. Therefore, to measure a small difference in temperature Beckmann’s thermometer is used. It is more sensitive than ordinary thermometer because one degree is further divided into hundred divisions.

            (e)        In summar the antifreeze solutions protect the radiator from boiling over.
Ans.    Water is used as a coolant in automobile radiation to decrease the temperature of the working engine. Pure water boils at 100oC. It is observed that solutions boil at higher temperatures than do pure liquid. So, an aqueous solution of antifreeze such as ethylene glycol is used in place of pure water in radiators because it raises the boiling point of pure water. Hence, in summer the antifreeze solution protects the liquid of the radiator to boil over.

            (f)         NaC1 and KNO3 are used to lower the melting point of ice.
Ans.    It is a common observation that the freezing point of solution is always lower than the freezing point of the pure solvent. The lowering in freezing point depends upon the number of solute particles (molecules/ ions).
            A mixture of NaC1 and KNO3 salts is used as a freezing mixture to lower melting points of ice. These salts dissociate in ice water. They split up into a number of ions each of which as a particle due to which the freezing point of water, i.e., the melting point of ice is lowered.
Q13.    What is Raoult’s law? Give is three statements. How this law can help us to understand the ideality of a solution?
Q14.    Give graphical explanation for elevation of boiling point of a solution. Describe one method to determine the boiling pointy elevation of a solution.
Q15.    Freezing points of solutions are depressed when non-volatile solutes are present in volatile solvents. Justify it. Plot a graph to elaborate your answer. Also give one method to record the depression of freezing point of a solution.
Q16.    Discuss the energetics of solution. Justify the heats of solutions as exothermic and endothermic properties.
Q17.    (a)        Calculate the molarity of glucose solution when 9 g of it are dissolved in 250 cm3 of solution.
Solution:
(a)                Mass of glucose C6H12O6          =9g
Molar mass of C6H12O6              =180g mol-1
            Vol. of solution           =250 cm3 =0.25dm3
            Molarity                      =?
Formula Used:            Molarity=x
                        Molarity   =x
                        Molarity   =0.2 mol dm-3 Answer
(b)        Calculate the mass of urea in 100 g of H2O in 0.3 molal solutions.
Solution:
                        Molality=0.3
            Mass of H2O     =100 g =0.1 kg
            Mass of urea (solute)      =?
            Molar mass of urea, NH2CONH2=60 g mol-1
Formula Used:            Molarity= x 
            0.3 mol kg-1          =x
            Mass of urea         = 0.3 mol kg-1 x 60 g mol-1x 0.1 kg
            Mass of urea       = 1.8 g Answer
(c)        Calculate the concentration of a solution in moles kg-1, which is obtained by mixing 250 g of 20% solution of NaC1 with 200 g of 40% solution of NaC1.
Solution:
            First solution:
100 g of NaC1 solution contains pure NaC1 =20g
250 g of NaC1 solution contains pure NaC1 =x250 g =80g
            Second solution:
100 g of solution contains pure NaC1                        =40g
200 g of NaC1 solution contains pure            =x200 g =80g
Total mass of solute NaC1                  =(50 + 80)g =130 g
Total mass of NaC1 solution              =(250 + 200) g =450 g
Mass of solvent =(450 – 130)g =320 g =0.32 kg
Formula used:
Molarity = x 
            Molality =x
            Molality=6.94 mol kg-1 Answer
Q18.    (a)        An aqueous solution of sucrose has been labeled as 1 molal. Find the mole fraction of the solute and the solvent.
Solution:
                                    Molality           =1
            Mass of solute, C12H22O11         =?
            Molar mass of  C12H22O11         =342 g mol-1
                        No. of kg of solvent    =1 kg
            Formula used:
Molarity = x 
Mass of solute, =molality x molar mass of solute x No. of Kg of solvent
                                    = 1 x 342 x 1
                                    =342 g
Now,   No. of moles of sucrose==1mol
            No. of moles of water  =  =55.56 mol
            Total no. of moles       =1 + 55.56 = 56.56 moles
            Mole fraction of sucrose ==0.076 Answer
            Mole fraction of water    ==0.9823 Answer
(b)        Your are provide with 80% H2SO4 having density 1.84 g cm-3. How much volume of this H2SO4sample is required to obtain one dm3 of 20% H2SO4 which has a density of 1.25 g cm-3.
Solution:
            First solution: % of H2SO4     =80
                                    Density of H2SO4       = 1.8 g cm-3
It means that:  1 cm3 of H2SO4 has mass        =1.8 g
            1000 cm3 H2SO4 has mass                  =1.8 x1000=1800g
Now, 100g of H2SO4 solution contains pure H2SO4 =80 g
1800 g of H2SO4 solution contains pure H2SO4 =x 1800 =1440 g
            No. of moles of H2SO4 =
                                                  ==14.7 mol
So, 14.7 moles of H2SO4 are present in 100 cm3 of concentrated H2SO4 solution, therefore , the molarity of solution is 14.7.
                        Molarity of conc. H2SO4 = 14.7 M
Second solution:        Percentage of dilute H2SO4 solution =20% (w/w)
            Density of dilute H2SO4 solution =1.25 g cm-3
It means that:  1 cm3 of H2SO4 has mass=1.25 g
            1000 cm3 of H2SO4 has mass       =1.25 x 1000   =1250 g
Now, 100 g dilute H2SO4 solution contains pure H2SO4 =20g
1250 g of dilute H2SO4 solution contains pure H2SO4 =x1250 =250 g
                        No. of moles of H2SO4 = =2.55 mol
So, 2.55 moles of H2SO4 are present in 1000 cm3 of dilute H2SO4 solution, therefore the molarity of dilute H2SO4 solution is 2.55 M.
Now, volume of conc. H2SO4 solution required to prepare dilute H2SO4 solution can be calculated by using the dilution formula:
                                    Conc. H2SO4               dilute H2SO4
                        M1V1               =          M2V2
                        14.7 xV1          =          2.55 x 1000cm3
                                            V1
                        V1=173.46 cm3 =173.5 cm3
Hence, volume of concentration H2SO4solution required to prepare dilute H2SO4
Solution =175.5 cm3 Answer
Q19.    250 cm3 of 0.2 molar K2SO4 solution is mixed with 250 cm3  of 0.2 molar KC1 solution. Calculate the molar concentration of K+ ions in the solution.
Solution:
                        K2SO42K++ SO                                                       
            0.2 M               2 x 0.2
                                        0.4
            Molarity of K+ ions=0.4M
            KC1K++ C1
            0.2 M           0.2M
            Molarity of K+ ions =0.2M
Total molarity of K+ ions=(0.4 +0.2) M=0.6M
Total volume of solution =250 cm3 +250 cm3=500cm3
Since after mixing the two solutions, the total volume becomes 500 cm3, so the concentration of K+ ions becomes half. So,
                        Molarity of K+ ions ==0.3 M Answer
Q20.    5 g of NaC1 are dissolved in 1000 g of water. The density of resulting  solution is 0.997 g cm-3. calculate molality, molarity and mole-fraction of this solution.
            Assume that the volume of the solution is equal to that of solvent.
Solution:
(i)                 Calculations for Molality
Mass of solute, NaC1 =5g
Molar mass of solute  =58.5 g mol-1
Mass of solvent , H2O =1000g =1kg
                        Molality=?
Formula used:
Molarity = x 
                        Molality =
                                    =0.0854 mol kg-1 = 0.0854 m Answer
(ii)        Calculations for molarity:
                        Mass of solute, NaC1 =5g
                        Molar mass of solute   =58.5 g mol-1
                        Mass of solvent, H2O   =1000 g = 1kg
                                    Density of solution=0.997 g cm3
            Now,               d     or         V=
            Vol. of solution ==1003 cm3
            Vol. of solution in dm3 ==1.003 cm3
            Molarity = x 
                           =
                            =0.0852 mol dm-3 = 0.0852 M Answer
(ii)               Calculations for mole fraction:
No. of moles of solute, NaC1 ==0.0855 mol
No. of moles of solvent H2O   ==55.556 mol
Total number of moles            =0.0855 + 55.556 =55.64 mol
Mole fraction of NaC1, X NaC1==0.00154 Answer
Mole fraction of water, XH2O               ==09984 Answer
Q21.    4.675 g of compound with empirical formula C3H3O were dissolved in 212.5 g of pure benzene. The freezing point of solution was found 1.02oC less than that of pure benzene. The molal freezing point constant of benzene is 5.1oC. calculate (i) the relative molar mass and (ii) the molecular formula of the compound.
Solution:
(i)                                                                 Mass of solute=4.675 g
Mass of solvent           =212.5 g
                        Kf        =5.1oC
                        Tf     =1.02oC
Molar mass of solute   =?
Formula used:          
Molar mass of solute   ==x 1000
                        Molar mass of solute ==
                        Molar mass of solute =110 g mol-1 Answer
(ii)        Empirical formula                   =C3 H3O
            Empirical formula mass =36+3+16=55
                                    n =
                                    n ==2
Molecular formula       =(Empirical formula)n
                                    =( C3 H3O)2
                                    = C3 H3O2 Answer
Q22.    The boiling point of a solution containing 0.2 g of a substance A in 20.0 g of ether (molar mass =74) is 0.17 K higher than that of pure ether. Calculate the molar mass of A. Molal boiling point constant of ether is 2.16 K.
Solution:
                                    Mass of A=0.2 g
                        Mass of solvent   =20 g
                                    Tb         =0.17 k
                                       Kb         =2.16K
            Molar mass of  solute, A=?
Formula used:           Molar mass of Ax x1000.
            Molar mass of A         =xx1000
                        Molar mass of A=127 g mol-1 Answer
Q23.    3 g of a non-volatile , non-electrolyte solute ‘X’ are dissolved in 50 g of ether (molar mass =74) at 293 K. The vapour pressure of ether falls from 442 torr to 426 torr under these conditions. Calculate. Calculate the molar mass of solute ‘X’.
Solution:
                                    Mass of solute, X =3g
                        Mass of solvent (ether)    =50 g
            Molar mass of solvent (ether)      =74
Vapour pressure of pure solvent, Po          =442 torr
            Vapour pressure of solution, p      =426 torr
                                                P=Po – p =442 – 426 =16 torr
                        Molar mass of solute, X    =?
Formula used:
Molar mass of solute, X =x x Molar mass of solvent
                                    =
                                    =122.66 g mol-1 Answer

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