Discussion Forum



1)

9-gram anhydrous oxalic acid (mol.wt.=90) was dissolved in 9.9 moles of water. If the vapour  pressure of pure water is p1° the vapour  pressure of the solution is 


A) $0.99p_1^0$

B) $0.1p_1^0$

C) $0.99p_1^0$

D) $1.1p_1^0$

Answer:

Option A

Explanation:

  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 (pw)

According  to Raoult's law= 

 $p_{w}=x_{w}p_w^o$

 Number of moles  of oxalic acid = $\frac{9}{90}$= 0.1 moles

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

 $\Rightarrow $   $p_{s}=p_{w}=$    $0.99p_1^0$