Discussion Forum

1)

How many grams of glucose are required to prepare an aqueous solution of glucose having a vapour pressure of 23.324 mm Hg at $25^{0} C$ in 100 g of water? The vapour pressure of pure water at $25^{0}C$ is 23.8 mmHg.

(Molar mass of glucose =180 g mol^-1) 

Answer:

Option A

Explanation:

Vapour pressure of pure solvent

 = $p^{0}$=23.8 mm of Hg

 Vapour pressure of solution  (p0) =23.324 mm of Hg

Thus, difference in vapour pressure($\triangle$p)

=23.8-23.324=0.476mm of Hg

$\because$        $\frac{\triangle p}{p^{0}}=\frac{w_{B}}{M_{B}} \times  \frac{M_{A}}{w_{A}}$   (for dilute solution)

where, M_A and w_A are molar mass and actual mass of solvent respectively

 M_B and w_B are molar mass and actual mass of solute respectively;

Given;

 M_A=18, w_A=100

w_B=M_B=180

$w_{B}=\frac{\triangle p \times M_{B} \times w_{A}}{p^{0} \times M_{A}}$

   =$=\frac{0.02 \times180\times100}{18}$       $\left[\frac{\triangle p}{p^{0}}=\frac{0.476}{23.8}=0.02\right]$ 

  =20.00

Hence, option (a) is the correct answer