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. $0.99p_1^{\circ }$
• B. $0.1p_1^{\circ }$
• C. $0.90p_1^{\circ }$
• D. $1.1p_1^{\circ }$

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

$p_w=x_w{p}_w^{\circ }$

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

$\therefore x_w=\frac{9.9}{9.9+0.1}=0.99$

$\Rightarrow p_s=p_w=0.99\times p_1^{\circ }$