5% solution of sugar cane (mol. wt = 342) is isotonic with 1 % solution of X under similar conditions. The mol. mass of X is

Question

5% solution of sugar cane (mol. wt = 342) is isotonic with 1 % solution of X under similar conditions. The mol. mass of X is (A) 136.2 (B) 68.4 (C) 34.2 (D) 171.2

Tardigrade Admin · Accepted Answer

π1 = π2 C1 = C2 (5 / 342/0.1) = (1/M/0.1) (5/342) = (1/M) ⇒ M = (342/5) = 68.4 gm/mol

Q. 5% solution of sugar cane (mol. wt = 342) is isotonic with 1 % solution of X under similar conditions. The mol. mass of X is

136.2

68.4

34.2

171.2

$π_{1} = π_{2}$
$C_{1} = C_{2} < b r / > \frac{5/342}{0.1} = \frac{1/ M}{0.1}$
$\frac{5}{342} = \frac{1}{M} \Rightarrow M = \frac{342}{5} = 68.4 g m / m o l$

Q. 5% solution of sugar cane (mol. wt = 342) is isotonic with 1 % solution of X under similar conditions. The mol. mass of X is

136.2

68.4

34.2

171.2

π1​=π2​ C1​=C2​<br/>0.15/342​=0.11/M​ 3425​=M1​⇒M=5342​=68.4gm/mol

$π_{1} = π_{2}$
$C_{1} = C_{2} < b r / > \frac{5/342}{0.1} = \frac{1/ M}{0.1}$
$\frac{5}{342} = \frac{1}{M} \Rightarrow M = \frac{342}{5} = 68.4 g m / m o l$