Question

Mole fraction of $C_3{H}_5(OH{)}_3$ in a solution of 36 g of water and 46 g of glycerine is:

• A. 0.46
• B. 0.36
• C. 0.2
• D. 0.4

Answer 1

The correct option is C 0.2

$C_3{H}_5{\left(OH\right)}_3\Rightarrow C_3{H}_8{O}_3$

$36g$ of $H_{2}O$, $46g$ of $\text{glycerine}$.

No. of moles of glycerine$=\frac{46}{92}$ (where 92 is M.M. of glycerine)

=0.5moles

no. of moles of water$=\frac{36}{18}$ (where 18 is M.M. of water)

=2moles

$\text{Mole fraction of glycerine}=\frac{M_G}{M_W+M_G}=\frac{0.5}{0.5+2}=0.2$

Hence, the correct option is $C$