A solution containing 10 g per dm 3 of urea (molecular mass = 60 g mol-1) is isotonic with a 5 % solution of a non-volatile solute. The molecular mass of this non-volatile solute is :

Question

A solution containing 10 g per dm 3 of urea (molecular mass = 60 g mol-1) is isotonic with a 5 % solution of a non-volatile solute. The molecular mass of this non-volatile solute is : (A) 200 g mol-1 (B) 300 g mol-1 (C) 400 g mol-1 (D) 500 g mol-1

Tardigrade Admin · Accepted Answer

For isotonic solution π1=π2 or C 1= C 2 (conc. in mol/lit.) (Urea solution) (unknown solution) (10/60)=(5 × 1000/ m w × 100) m w =300 gm mol -1

Q. A solution containing $10 g$ per $d m^{3}$ of urea (molecular mass = $60 g m o l^{- 1})$ is isotonic with a $5%$ solution of a non-volatile solute. The molecular mass of this non-volatile solute is :

$200 g m o l^{-} 1$

$300 g m o l^{- 1}$

$400 g m o l^{- 1}$

$500 g m o l^{- 1}$

For isotonic solution $π_{1} = π_{2}$

or $C_{1} = C_{2}$ (conc. in mol/lit.)

(Urea solution) (unknown solution)

$\frac{10}{60} = \frac{5 \times 1000}{m _{w} \times 100}$

$m_{w} = 300 g m m o l^{- 1}$

Q. A solution containing 10g per dm3 of urea (molecular mass = 60gmol−1) is isotonic with a 5% solution of a non-volatile solute. The molecular mass of this non-volatile solute is :

200gmol−1

300gmol−1

400gmol−1

500gmol−1