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A solution containing 10 g per dm$^3$ of urea (molecular mass = 60 g mol$^{-1}$) is isotonic with a 5% solution of a nonvolatile solute. The molecular mass of this nonvolatile solute is :
A
250 g mol$^{-1}$
B
300 g mol$^{-1}$
C
350 g mol$^{-1}$
D
200 g mol$^{-1}$
Detailed Solution
Isotonic solutions have the same osmotic pressure: $\pi_1 = \pi_2$
Since $\pi = CRT$, at the same temperature $C_1 = C_2$ (concentrations in mol per litre), both solutes being non-electrolytes.
Urea solution: 10 g per dm$^3$, so $C_1 = \dfrac{10}{60}$ mol L$^{-1}$
Unknown solution: 5 percent means 5 g in 100 mL, i.e. 50 g per litre, so $C_2 = \dfrac{5 \times 1000}{M_w \times 100}$ mol L$^{-1}$
$\dfrac{10}{60} = \dfrac{5 \times 1000}{M_w \times 100}$
$\dfrac{1}{6} = \dfrac{50}{M_w}$
$M_w = 300$ g mol$^{-1}$
Since $\pi = CRT$, at the same temperature $C_1 = C_2$ (concentrations in mol per litre), both solutes being non-electrolytes.
Urea solution: 10 g per dm$^3$, so $C_1 = \dfrac{10}{60}$ mol L$^{-1}$
Unknown solution: 5 percent means 5 g in 100 mL, i.e. 50 g per litre, so $C_2 = \dfrac{5 \times 1000}{M_w \times 100}$ mol L$^{-1}$
$\dfrac{10}{60} = \dfrac{5 \times 1000}{M_w \times 100}$
$\dfrac{1}{6} = \dfrac{50}{M_w}$
$M_w = 300$ g mol$^{-1}$