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. $684$
• B. $171$
• C. $100$
• D. $342$

Answer 1

The correct option is B $171$

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

or $\stackrel{{\pi }_{\left(sucrose\right)}={\pi }_{\left(organicsubstance\right)}}{CR{T}_{\left(sucrose\right)}=CR{T}_{\left(organicsubstance\right)}}$
or ${[\frac{w}{M\times V}\cdot RT]}_{\left(sucrose\right)}={[\frac{w}{M\times V}\cdot RT]}_{\left(organicsubstance\right)}$
or $\frac{6}{342}=\frac{3}{M_{\left(organicsubstance\right)}}$
where, $w_{\left(sucrose\right)}=6$
$w_{\left(organicsubstance\right)}=3$
$M_{\left(sucrose\right)}=3$
$M_{\left(organicsubstance\right)}=?$
Hence, $M_{\left(organicsubstance\right)}=\frac{3\times 342}{6}$

$\therefore$ Molecular mass of unknown organic substance
$=171$.