Question

the vapour pressure of ccl4at 25 celcius is 143 mm hg . of 0.5 gm of a non volatile solute (mol. weight=65) is dissolved in 100 g ccl4, the vapour pressure of the solution will be

Answer 1

weight of non volatile solute , w = 0.5gm

molecular weight of solute , W = 65gm/mol

so, number of mole of solute , n = w/W

= 0.5/65 = 1/130 = 0.0077

weight of CCl4 , m = 100g

molecular weight of of CCl4 , M = (12 + 4 × 35.5) = 154g/mol

so, number of mole of CCl4 , N = m/M

= 100/154 = 0.65

here it is clear that n << N so, (n + N) ≈ N

so, mole fraction of solute = n/(n + N) = n/N

from lowering of vapor pressure,

$\frac{(P_0-P)}{P_0}$ = mole fraction

given, vapor pressure of pure CCl4 , $P_0$ = 143mm Hg

now, (143 - P)/143 = 0.0077/0.65

or, 143 × 0.0077 = 0.65(143 - P)

or, 1.011 = 92.95 - 0.65P

or, P = 141.44mm Hg

hence, answer is 141.44mm Hg

Answer 2

Answer:

141.44mm Hg

Explanation:

Vapour pressure of ccl4 = P0 = 143 mm hg (Given)

Temperature of ccl4 = 25°C  (Given)

Weight of CCl4 = 100g (Given)

Weight of the non volatile solute , w = 0.5gm (Given)

Molecular weight of solute , W = 65gm/mol (Given)

Thus, the number of mole of solute - n = w/W

= 0.5/65

= 1/130

= 0.0077

Molecular weight of of CCl4 -

= 12 + 4 × 35.5

= 154g/mol

Thus, the number of mole of CCl4 - N = m/M

= 100/154

= 0.65

Equation of fraction of mole = P0-P/P0

=  (143 - P)/143

= 0.0077/0.65  

or, 143 × 0.0077 = 0.65(143 - P)

= 1.011 = 92.95 - 0.65P

= 141.44mm Hg

Thus, the vapour pressure of the solution will be 141.44mm Hg.