Solved Example
Ex.1 The molarity of 20% (W/W) solution of 55.5
sulphuric acid is 2.55 M. The density of the = = 55.5 M
1
solution is : (Ans. A)
–3 Ex.4 Calculate the quantity of sodium carbonate
(A) 1.25 g cm (B) 0.125 g L–1 (anhydrous) required to prepare 250 ml of
(C) 2.55 g cm–3 (D) unpredictable 0.1 M solution-
100 (A) 2.65 gram (B) 4.95 gram
Sol. Volume of 100 g of solution = ml
d (C) 6.25 gram (D) None (Ans. A)
20 d 1000 Sol. We know that
M=
100 98 Molarity =
W
2.55 100 98 MV
or d= = 1.249 1.25 where;
20 1000
W = Mass of Na2CO3 in gram
Ex.2 The density of a solution containing 13% by M = Molecular mass of Na2CO3 in grams =106
mass of sulphuric acid is 1.09 g/mL. Calculate 250
the molarity and normality of the solution- V =Volume of solution in litres = = 0.25
1000
(A) 1.445 M (B) 14.45 M
(C) 144.5 M (D) 0.1445 M 1
Molarity =
(Ans. A) 10
100 1 W
Sol. Volume of 100 gram of the solution = Hence , = =
d 10 106 0.25
100 100 106 0.25
= mL = litre or W= = 2.65 gram
1.09 1.09 1000 10
1 Ex.5 The freezing point of a 0.08 molal aqueous
= litre
1.09 10 solution of NaHSO4 is – 0.372°C. The
Number of moles of H2SO4 in 100 gram of the dissociation constant for the reaction
HSO4– H+ + SO42– is
13 –2
solution = (A) 4 × 10 (B) 8 × 10–2
98 –2
(C) 2 × 10 (D) 1.86 × 10–2
No. of moles of H 2SO4
Molarity = (Ans. A)
Volume of solution in litre Sol. Tf = Kf × m
13 1.09 10 Tf 0.372
= × = 1.445 M m= 0.2
98 1 Kf 1.86
This means that total molal conc. of all particles
Ex.3 Calculate the molarity of pure water is 0.2.
(d = 1gm/mL) NaHSO4 Na+ + HSO4–
(A) 555 M (B) 5.55 M 0.08 0.08
(C) 55.5 M (D) None (Ans. C) HSO4– H+ + SO42–
Sol. Consider 1000 mL of water
(0.08 – x) x x
Mass of 1000 mL of water
The net particle concentration
= 1000 × 1 = 1000 gram
0.08 + 0.08 – x + x + x = 0.2
1000 or x = 0.04
Number of moles of water = = 55.5
18
[H ][SO 24 – ]
No. of moles of water Ka = = 0.04
Molarity = [HSO 4– ]
Volume in litre
SOLUTION & COLLIGATIVE PROPERTIES 54
54
,Ex.6 Find the molality of H2SO4 solution whose Moles of solute
Molality of solution =
specific gravity is 1.98 g ml-1 and 95% by wt of solvent in kg
volume H2SO4
5
(A) 7.412 (B) 8.412
(C) 9.412 (D) 10.412 (Ans. C) = 60 789 = 0.1056
1000
Sol. H2SO4 is 95% by volume
Ex.9 Calculate the molarity and normality of a
wt. of H2SO4 = 95g solution containing 0.5 gm of NaOH dissolved
Vol of solution = 100ml in 500 ml. solution-
(A) 0.0025 M, 0.025 N
95
Moles of H2SO4 = , and weight of (B) 0.025 M, 0.025 N
98 (C) 0.25 M, 0.25 N
solution = 100 × 1.98 = 198 g (D) 0.025 M, 0.0025 N (Ans. B)
Weight of water = 198 – 95 = 103 g Sol. Wt. of NaOH dissolved = 0.5 gm
95 1000 Vol. of NaOH solution = 500 ml
Molality = = 9.412
98 103 Calculation of molarity
Hence molality of H2SO4 solution is 9.412 0.5
0.5 g of NaOH = moles of NaOH
40
Ex.7 Calculate molality of 1 litre solution of [ Mol. wt of NaOH = 40]
93% (w/v) H2SO4. The density of solution is = 0.0125 moles
Thus 500 ml of the solution contain
1.84 gm ml–1
NaOH = 0.0125 moles
(A) 9.42 (B) 10.42
1000 ml of the solution contain
(C) 11.42 (D) 12.42 (Ans. B)
0.0125
Sol. Given H2SO4 is 93% by volume = × 1000
500
wt. of H2SO4 = 93g = 0.025 M
Volume of solution = 100ml Hence molarity of the solution = 0.025 M
weight of solution = 100 × 1.84 gm Calculation of normality
= 184 gm Since NaOH is monoacidic ;
wt. of water = 184 – 93 = 91 gm Eq. wt. of NaOH = Mol. wt. of NaOH = 40
0.5
Moles 0.5 gm of NaOH = gm equivalents
Molality = 40
wt. of water in kg
= 0.0125 gm equivalents
93 1000 Thus 500 ml of the solution contain
= = 10.42
98 91 NaOH = 0.0125 gm equ.
1000 ml of the solution contain
Ex.8 Suppose 5gm of CH3COOH is dissolved in one 0.0125
= × 1000 = 0.025
litre of Ethanol. Assume no reaction between 500
them. Calculate molality of resulting solution if Hence normality of the solution = 0.025 N
density of Ethanol is 0.789 gm/ml. Ex.10 Calculate the molality and mole fraction of the
(A) 0.0856 (B) 0.0956 solute in aqueous solution containing 3.0 gm of
(C) 0.1056 (D) 0.1156 (Ans. C) urea per 250 gm of water (Mol. wt. of urea =
Sol. Wt . of CH3COOH dissolved = 5g 60).
(A) 0.2 m, 0.00357 (B) 0.4 m, 0.00357
5 (C) 0.5 m, 0.00357 (D) 0.7m, 0.00357
Eq. of CH3COOH dissolved =
60 (Ans. A)
Volume of ethanol = 1 litre = 1000ml. Sol. Wt. of solute (urea) dissolved = 3.0 gm
Weight of ethanol = 1000 × 0.789 = 789g Wt. of the solvent (water) = 250 gm
Mol. wt. of the solute = 60
SOLUTION & COLLIGATIVE PROPERTIES 55
55
, 3 .0 (A) 30% (B) 50%
3.0 gm of the solute = moles = 0.05 moles (C) 70% (D) 75% (Ans. A)
60
Sol. Total mass of solution = (15 + 35) gram = 50
Thus 250 gm of the solvent contain = 0.05
gram
moles of solute
mass percentage of methyl alcohol
1000 gm of the solvent contain = Mass of methyl alcohol
0.05 1000 = × 100
= 0.2 moles Mass of solution
250
15
Hence molality of the solution = 0.2 m = × 100 = 30%
In short, 50
Molality = No. of moles of solute/1000 g of Ex.13 Calculate the masses of cane sugar and water
solvent required to prepare 250 gram of 25% cane sugar
solution-
Molality = × 1000 = 0.2 m (A) 187.5 gram, 62.5 gram
250
(B) 62.5 gram, 187.5 gram
Calculation of mole fraction
(C) 162.5 gram, 87.5 gram
3.0 gm of solute = 3/60 moles = 0.05 moles
(D) None of these (Ans. B)
250
250 gm of water = moles Sol. Mass percentage of sugar = 25
18 We know that
= 13.94 moles Mass of solute
Mole fraction of the solute Mass percentage = × 100
Mass of solution
0.05 0.05
= = Mass of cane sugar
0.05 13.94 13.99 So, 25 = × 100
= 0.00357 250
25 250
Ex.11 A solution has 25% of water, 25% ethanol and or Mass of cane sugar =
100
50% acetic acid by mass. Calculate the mole = 62.5 gram
fraction of each component.
Mass of water = (250 – 62.5) = 187.5 gram
(A) 0.50, 0.3, 0.19 (B) 0.19, 0.3, 0.50
(C) 0.3, 0.19, 0.50 (D) 0.50, 0.19, 0.3
Ex.14 Calculate normality of the mixture obtained by
(Ans. D)
Sol. Since 18 g of water = 1 mole mixing 100ml of 0.1N HCl and 50ml of 0.25N
NaOH solution.
25
25 g of water = = 1.38 mole (A) 0.0467 N (B) 0.0367 N
18 (C) 0.0267 N (D) 0.0167 N
Similarly, 46 g of ethanol = 1 mole (Ans. D)
25 Sol. Meq. of HCl =100 × 0.1 = 10
25 g of ethanol = = 0.55 moles
46 Meq. of NaOH = 50 × 0.25 = 12.5
Again, 60 g of acetic acid = 1 mole HCl and NaOH neutralize each other with
50 equal eq.
50 g of acetic acid = = 0.83 mole
60 Meq. of NaOH left = 12.5 – 10 = 2.5
Mole fraction of water Volume of new solution = 100 + 50 = 150 ml.
1.38 2 .5
= = 0.50 NNaOH left = = 0.0167 N
1.38 0.55 0.83 150
Similarly, Mole fraction of ethanol Hence normality of the mixture obtained is
0.55 0.0167 N
= = 0.19
1.38 0.55 0.83
Mole fraction of acetic acid Ex.15 300 ml 0.1 M HCl and 200 ml of 0.03M H2SO4
0.83 are mixed. Calculate the normality of the
= = 0.3 resulting mixture-
1.38 0.55 0.83
(A) 0.084 N (B) 0.84 N
Ex.12 15 gram of methyl alcohol is dissolved in 35 (C) 2.04 N (D) 2.84 N (Ans.A)
gram of water. What is the mass percentage of Sol. For HCl For H2SO4
methyl alcohol in solution ?
SOLUTION & COLLIGATIVE PROPERTIES 56
56
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller ramanrathore. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $8.99. You're not tied to anything after your purchase.
