Question

A $1$% aqueous solution ($w/v$) of a certain substance is isotonic with a $3$% solution of dextrose i.e. glucose (molar mass $180$) at a given temperature. The molar mass of the substance is:

• A. $60$
• B. $120$
• C. $180$
• D. $360$

Answer 1

The correct option is A $60$

$1$% aqueous solution means $1$ g solute in $100$ mL solution.
So, $M_1=\frac{wt.}{Mol.wt.}\times \frac{1000}{V\left(ml\right)}$
$M_1=\frac{1}{(Mol.wt.{)}_1}\times \frac{1000}{100}$
Now, for $3$% solution of dextrose, $3$ g solute in $100$ g solution
$M_2=\frac{3}{180}\times \frac{1000}{100}$
Both the solution are isotonic $\Rightarrow$ same osmosis pressure
${\sqcap }_1={\sqcap }_2$
$\Rightarrow M_{1}RT=M_{2}RT$
$\Rightarrow \frac{1}{(Mol.wt.{)}_1}\times \frac{1000}{100}=\frac{3}{180}\times \frac{1000}{100}$
$(Mol.wt.{)}_1=\frac{180}{3}=60$