Question

A $5\%\left(w/V\right)$ solution of cane sugar is isotonic with $0.878\%\left(w/V\right)$ solution of urea. Calculate the molecular mass of urea if the molecular mass of cane sugar is $342{\text{g mol}}^{-1}$.

• A. $60.05{\text{g mol}}^{-1}$
• B. $120.1{\text{g mol}}^{-1}$
• C. $45.5{\text{g mol}}^{-1}$
• D. $76.1{\text{g mol}}^{-1}$

Answer 1

The correct option is A $60.05{\text{g mol}}^{-1}$

Isotonic solution:
Two solutions having the same osmotic pressure across a semipermeable membrane is referred to as an isotonic solution. It has the same osmolarity (solute concentration), as another solution.
$5\%\left(w/V\right)$ cane sugar solution means, $5g$ of cane sugar is there in $100\text{mL}$ of solution
$0.878\%\left(w/V\right)$ urea solution means, $0.878g$ of urea is there in $100\text{mL}$ of solution
$\text{Molar concentration of cane sugar}=\frac{w_1}{m_1\times V_1}=\frac{5}{342\times 0.1}$
and
$\text{Molar concentration of urea}=\frac{w_2}{m_2{V}_2}=\frac{0.878}{m_2\times 0.1}$
For isotonic solutions,

$\frac{w_1}{m_1{V}_1}=\frac{w_2}{m_2{V}_2}\frac{5}{342\times 0.1}=\frac{0.878}{m_2\times 0.1}{m}_2=\frac{0.878\times 342}{5}=60.05{\text{g mol}}^{-1}$