Question

10% solution of urea is isotonic with 6% solution of nonvolatile solute ‘x’. What will be the atomic mass of solute ‘x’.

• A. $6gmo{l}^{-1}$
• B. $60gmo{l}^{-1}$
• C. $36gmo{l}^{-1}$
• D. $32gmo{l}^{-1}$

Answer 1

Answer: (C)

Given value,
Solution of urea $=10\%$
Solution of non-volatile solute ${}^{\prime }x’=6\%$
$\because$ Isotonic solution has the same osmotic pressure and hence the same concentration in the solution.
We know that, osmotic pressure
$(\pi )=CRT$
here, $R$ and $T$ are constant
$\because {\pi }_1={\pi }_2$
$\therefore C_{1}R{T}_1=C_{2}R{T}_2$
$10\%$ solution of urea $=6\%$ of solution of ${}^{\prime }x’$
$\frac{10}{100\times 60}=\frac{6}{100\times x}$ (Molecular weight of urea $=60$)
$x=6\times 6=36gmo{l}^{-1}$
Hence, the atomic mass of solute ${}^{\prime }{x}^{\prime }$ is $36gmo{l}^{-1}$.