Question

$5$% solution of cane sugar is isotonic with $0.877$% of $X$. The molecular weight of substance $X$ is:

• A. $126.98$
• B. $119.96$
• C. $95.5$
• D. $59.98$

Answer 1

The correct option is D $59.98$

Let M g/mol be the molecular weight of substance X. The molecular weight of cane sugar is 342 g/mol.
Isotonic solutions have same osmotic pressure.
The osmotic pressure is given by the expression, $\pi =CRT$.
${\pi }_{\text{cane sugar}}={\pi }_{\text{X}}$
$C_{\text{cane sugar}}RT=C_{\text{X}}RT$
$C_{\text{cane sugar}}=C_{\text{X}}$
Here, concentration is in moles per litre. But we are given the concentration in weight / volume percent.
$C_{\text{cane sugar}}=C_{\text{X}}$

$(\frac{5}{342}{)}_{\text{cane sugar}}=(\frac{0.877}{M}{)}_{\text{X}}$

$M=0.877\times \frac{342}{5}$

$M=59.98$ g/mol.