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**Example ****14.** **Calculate normality of the mixture obtained by mixing 100ml of 0.1N HCl and 50ml of 0.25N NaOH solution.****(a) 0.0467 N ****(b) 0.0367 N ****(c) 0.0267 N ****(d) 0.0167 N****Ans. **(d)**Solution.**

Molal eq. of HCl = 100 Ã— 0.1 = 10

Molal eq. of NaOH = 50 Ã— 0.25 = 12.5

HCl and NaOH neutralize each other with equal eq.

Eq. of NaOH left = 12.5 - 10 = 2.5

Volume of new solution = 100 + 50 = 150 ml.

Hence normality of the mixture obtained is **0.0167 N ****Example ****15.** **300 ml 0.1 M HCl and 200 ml of 0.03M H _{2}SO_{4} are mixed. Calculate the normality of the resulting mixture-**

For HCl For H

V

N

= 0.1 Ã— 1 = 0.1 = 0.03 Ã— 2 = 0.06

Normality of the mixture

=** 0.084 N ****Example ****16. In what ratio should a 6.5 N HNO _{3} be diluted with water to get 3.5 N HNO_{3}?**

N

6.5 V

6.5 V

3 V

Amount of H_{2}SO_{4} per litre (strength) = Normality Ã— Eq. wt. = Ã— 49 = 7 g/litre

Amount in 150 ml

**(ii)** Molecular wt. of

NaHCO_{3} = 23 1 12 48 = 84

Amount of NaHCO_{3} required to produce 1000 c.c. of one molar solution = 84 g

Amount present per litre in 0.2 M solution = 84 Ã— 0.2 = 16.8 g

Amount present in 250 c.c.

= 4.2 g**(iii)** Equivalent weight of

= 53

Amount of Na_{2}CO_{3} = Normality Ã— Eq. wt. = Ã— 53 = 5.3 g/litre

Amount present in 400 c.c.

= 2.12 g**(iv) **We know that 1 molal solution of a substance contains 1000 g of solvent.

Wt. of KOH in 1052 g of 1 m KOH solution = 1052 - 1000 = 52 g**Example ****18.** **How many kilograms of wet NaOH containing 12% water are required to prepare 60 litres of 0.50 N solution ?****(a) 1.36 kg ****(b) 1.50 kg ****(c) 2.40 gm ****(d) 3.16 kg****Ans. **(a)**Solution.**

One litre of 0.50 N NaOH contains = 0.50 Ã— 40g = 20 g = 0.020 kg

60 litres of 0.50 N NaOH contain

= 0.020 Ã— 60 kg = 1.20 kg NaOH

Since the given NaOH contains 12% water, the amount of pure NaOH in 100 kg of the given NaOH = 100 - 12 = 88 kg

Thus 88 kg of pure NaOH is present in 100 kg wet NaOH

1.20 kg of pure NaOH is present in

= **1.36 kg wet NaOH ****Example ****19.** **Calculate the vapour pressure of a solution at 100 ^{0}C containing 3g of cane sugar in 33g of water. (At wt. C = 12 , H = 1 , O = 16)**

Vapour pressure of pure water (solvent) at 1000C, pÂ° = 760 mm.

Vapour pressure of solution, p = ?

Wt. of solvent, W = 33g

Wt. of solute, w = 3g

Mol. wt. of water (H

Mol. wt. of sugar (C

m = (12 Ã— 12) + (22 Ã— 1) + (11 Ã— 16) = 342

According to Raoult's law,

(pÂº

= 760 - 3.19 = 756.90 mm

Here it is given that

p = 2.5 atm, T = 24 273 = 297K, S = 0.0821 lit. atm. deg

We know that p = CST

or

= 0.1025 moles/litre**Example ****21.** **Twenty grams of a substance were dissolved in 500 ml. of water and the osmotic pressure of the solution was found to be 600 mm of mercury at 15ÂºC. Determine the molecular weight of the substance-****(a) 1120 ****(b) 1198 ****(c) 1200 ****(d) None of these****Ans.** (c)**Solution.**

Here it is given that

w = 20 gm ; V = 500 ml.

= 500/1000 = 0.5 litre

p = 600 mm = 600/760 atm;

T = 15 273 = 288^{0}A

m = ?

According to Van't Hoff equation,

pV = nST pV

= 1198**Example ****22.** **Blood plasma has the following composition (milli-equivalents per litre). Calculate its osmotic pressure at 37 ^{0}C.**

Since for calculating osmotic pressure we require millimoles/litre therefore

Na

Mg

HCO

SO

others = 1.0

Total = 294.18 millimoles/litre = 294.18/1000

= 0.294 moles/litre

Now since p = CST

= 0.294 Ã— 0.0821 Ã— .310 = 7.47 atm

Here it is given that

w = 0.15 g, DT

W = 15g K

m = ?

Substituting values in the expression,

= 100

Wt. of solute, w = 0.450 g

Wt. of solvent, W = 22.5 g

Mol. wt of solute, m = 60

Molal elevation constant K_{b} = ?

Boiling point elevation, DT_{b} = 0.170^{0}C

Substituting these values in the equation

=

= 0.51^{0}C**Example ****25. Calculate the boiling point of a solution containing 0.45g of camphor (mol. wt. 152) dissolved in 35.4g of acetone (b.p. 56.3 ^{0}C); K_{b} per 100 gm of acetone is 17.2^{0}C.**

Here it is given that

w = 0.45 g, W = 35.4, m = 152,

Kb = 17.2 per 100gm

Now we know that Î”T

(Note that this is expression when K_{b} is given per 100g of the solvent)

Substituting the values in the above expression.

= 0.146^{0}C

Now we know that

B.P. of solution (T) - B.P. of solvent (T_{0}) = DT

B.P. of solution (T) = B.P. of solvent(T_{0}) DT

Hence B.P. of solution = 56.3 0.146

= 56.446^{0}C**Example ****26.** **The freezing point of 0.2 molal K _{2}SO_{4} is _1.1ÂºC. Calculate Van't Haff factor and percentage degree of dissociation of K_{2}SO_{4}. K_{f} for water is 1.86Âº**

Î”T

We know that,

Î”T

1.1 = i Ã— 1.86 Ã— 0.2

But we know

i = 1 (n - 1) a

2.95 = 1 (3 - 1) a = 1 2a

a = 0.975

Van't Haff factor (i) = 2.95

Degree of dissociation = 0.975

Percentage degree of dissociation =

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