Question

$9$ gram anhydrous oxalic acid (mol. wt. = $90$) was dissolved in $9.9$ moles of water. If vapour pressure of pure water is $p_{1}^\circ 1$ the vapour pressure of solution is

• A. $0.99 p_{1}^{\circ}$
• B. $0.1 p_{1}^{\circ}$
• C. $0.90 p_{1}^{\circ}$
• D. $1.1 p_{1}^{\circ}$

Answer 1

Answer: (A)

The total vapour pressure of a solution in this case only depends on vapour pressure of water as anhydrous oxalic acid is a non-volatile compound

$\therefore $ Vapour pressure of solution = vapour pressure of water $\left(p_{w}\right)$

According to Raoult's law

$p_{w}=x_{w} p^{\circ}_{w}$

Number of moles of oxalic acid $=\frac{9}{90}=01$ moles

$\therefore x_{w}=\frac{9.9}{9.9+0.1}=0.99$

$\Rightarrow p_{s}=p_{w}=0.99 \times p^{\circ}_{1}$