Question

A solution containing $8.6 \,g \,L ^{-1}$ of urea is isotonic with a $5 \%$ solution of unknown solute. The molar mass of the solute will be

• A. $348.9 \,g \,mol ^{-1}$
• B. $174.5 \,g\, mol ^{-1}$
• C. $87.3 \,g \,mol ^{-1}$
• D. $34.89\, g\, mol ^{-1}$

Answer 1

Answer: (A)

Urea $\rightarrow 8.6 \,g / L =$ i.e., $\left(\frac{ w }{\text { volume }}\right)$
Molar mass $= NH _{2} CONH _{2} $
$\Rightarrow 14 \times 2+4+12+16 $
$\Rightarrow 60$
isotonic $\rightarrow\left[\pi_{1}=\pi_{2}\right]$
$C_{1} S T=C_{2} S T$
$C_{1}=C_{2}$
$\frac{W}{M_{w} \times V}=\frac{W}{M_{w} \times V}$
$\frac{8.6}{60} \Rightarrow \frac{5 \times 1000}{M_{w} \times 100}$
$M_{w}=\frac{5 \times 60 \times 10}{8.6}$
$=348.8 \,g / mol$
Errata : $5 \%$ solution of unknown solute but it must be written $5 \%$ weight/volume means $5\, g$ of solute present in $100\, mL$ of solution