Question

Calculate the vapour pressure of an aqueous solution of $1.0$ molal glucose solution at ${100}^{o}C.$

• A. $746.32\text{mm Hg}$
• B. $760\text{mm Hg}$
• C. $13.68\text{mm Hg}$
• D. $136.8\text{mm Hg}$

Answer 1

The correct option is A $746.32\text{mm Hg}$

By using the formula,
$\text{Molality}=\frac{w}{M\times W}\times 1000$
Where,
$\text{w = mass of solute in grams}\text{W = mass of solvent in grams}\text{M= molar mass of solute}\text{Substituting the values, we get}1.0=\frac{w}{M\times W}\times 1000$
$\Rightarrow \frac{w}{M\times W}=\frac{1.0}{1000}=0.001$
Applying Raoult's law for dilute solution,
$\frac{p_0-p_s}{p_0}=\frac{w}{M\times W}\times M_A(M_A=18g/mol)$
$\frac{760-p_s}{760}=0.001\times 18(p_0=\text{760 mm}at{100}^{o}C)$
$\Rightarrow p_s=760-760\times 0.001\times 18$
$=760-13.68$
$=746.32\text{mm Hg}$