Question

$A\, 6\%$ solution of sucrose $C_{22}H_{22} O_{11}$ is isotonic with $3\%$ solution of an unknown organic substance. The molecular weight of unknown organic substance will be:

• A. 342
• B. 684
• C. 171
• D. 100

Answer 1

Answer: (C)

171

Answer 2

$\because 6 \%$ sucrose solution is isotonic with $3 \%$ solution of unknown organic substance, thus,

$\pi_{\text {(sucrose) }}=\pi_{\text {(organic substance) }}$

or $C R T_{\text {(sucrose) }}=C R T_{\text {(organic substance) }}$

or $\left[\frac{W}{M \times V} \cdot R T\right]_{\text {(sucrose) }}=\left[\frac{W}{M \times V} \cdot R T\right]_{\text {(organic sunstance) }}$

or $\frac{6}{342}=\frac{3}{M_{\text {(organic substance) }}}$

where, $W_{\text {(sucrose) }} =6$
$W_{\text {(organic substance) }} =3$
$M_{\text {(sucrose) }} =342$
$M_{\text {(organic substance) }}$=?

Hence, $M_{\text {(organic substance) }} =\frac{3 \times 342}{6}$
$\therefore$ Molecular mass of unknown organic substance $=171$

The formula of sucrose in the question is given $C _{22} H _{22} O _{11}$, which is incorrect. The correct formula is $C _{12} H _{22} O _{11}$.