Question

A solution containing $1.8\text{g}$ of a non-electrolyte (empirical formula $C{H}_{2}O$) in $40\text{g}$ of water is observed to freeze at $-0.465{}^{\circ }\text{C}$. The molecular formula of the compound is:
($K_f$ of water $=1.86{\text{kg K mol}}^{-1})$

• A. $C_3{H}_6{O}_3$
  
• B. $C_4{H}_8{O}_4$
  
• C. $C_6{H}_{12}{O}_6$
  
• D. $C_2{H}_4{O}_2$
  

Answer 1

The correct option is C $C_6{H}_{12}{O}_6$


We know,
$\Delta T_f=K_f\times \text{molality}$
$\Delta T_f=\text{Depression of freezing point}=-0.465{}^{\circ }\text{C}$
$K_f=1.86{\text{kg K mol}}^{-1}$

$\therefore 0.465=1.86\times \frac{1.8}{M\times 40}\times 1000$
$\Rightarrow M=\frac{1.86\times 1.8\times 1000}{0.465\times 40}=180{\text{g mol}}^{-1}$

Empircal formula given = $C{H}_{2}O$
Let, molecular formula = $(C{H}_{2}O{)}_n$
Empirical formula mass $=30\text{u}$
Molecular formula mass $=180\text{u}$
$\therefore n=\frac{\text{Molecular formula mass}}{\text{Empirical formula mass}}=\frac{180}{30}=6$
So, molecular formula = $C_6{H}_{12}{O}_6$.