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NEET CHEMISTRYEasy

A solution containing 10 g/dm310 \text{ g/dm}^3 of urea (molecular mass =60 g mol1= 60 \text{ g mol}^{-1}) is isotonic with a 5%5 \% solution of a non-volatile solute. The molecular mass of this non-volatile solute is:

A

25 g mol⁻¹

B

300 g mol⁻¹

C

350 g mol⁻¹

D

200 g mol⁻¹

Step-by-Step Solution

Two solutions are said to be isotonic if they have the same osmotic pressure (π\pi) at a given temperature . The osmotic pressure is given by the formula π=CRT\pi = CRT, where CC is the molar concentration .

For isotonic solutions: π1=π2    C1RT=C2RT    C1=C2\pi_1 = \pi_2 \implies C_1RT = C_2RT \implies C_1 = C_2

  1. Calculate Concentration of Urea (C1C_1): Concentration given = 10 g/dm3=10 g/L10 \text{ g/dm}^3 = 10 \text{ g/L}. Molar mass of urea = 60 g mol160 \text{ g mol}^{-1}.
  • C1=1060 mol L1C_1 = \frac{10}{60} \text{ mol L}^{-1}.
  1. Calculate Concentration of Unknown Solute (C2C_2): Concentration given = 5%5 \%. In the context of osmotic pressure problems where density is not provided, this is interpreted as mass/volume percentage (w/vw/v), meaning 5 g5 \text{ g} solute in 100 mL100 \text{ mL} solution. Concentration in g/L = 5 g×1000 mL100 mL=50 g/L5 \text{ g} \times \frac{1000 \text{ mL}}{100 \text{ mL}} = 50 \text{ g/L}. Let the molecular mass of the unknown solute be M2M_2. C2=50M2 mol L1C_2 = \frac{50}{M_2} \text{ mol L}^{-1}.

  2. Equate and Solve: 1060=50M2\frac{10}{60} = \frac{50}{M_2} M2=50×6010=300 g mol1M_2 = \frac{50 \times 60}{10} = 300 \text{ g mol}^{-1}.

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