A 5.25 % solution of a substance is isotonic with 1.5 % solution of urea (molar mass =60gmol- 1 ) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0gcm- 3 , molar mass of the substance will be :

Question

A 5.25 % solution of a substance is isotonic with 1.5 % solution of urea (molar mass =60gmol- 1 ) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0gcm- 3 , molar mass of the substance will be : (A) 90.0gmol- 1 (B) 11.50gmol- 1 (C) 105.0gmol- 1 (D) 210.0gmol- 1

Tardigrade Admin · Accepted Answer

π 1=π 2 [.C].sol=[.C]. texturea (5 . 25/M)=(1 . 5/60) ⇒ M=(5 . 25 × 60/1 . 5) =210

Q. A $5.25%$ solution of a substance is isotonic with $1.5%$ solution of urea (molar mass $= 60 g m o l^{- 1}$ ) in the same solvent. If the densities of both the solutions are assumed to be equal to $1.0 g c m^{- 3}$ , molar mass of the substance will be :

$90.0 g m o l^{- 1}$

$11.50 g m o l^{- 1}$

$105.0 g m o l^{- 1}$

$210.0 g m o l^{- 1}$

$π_{1} = π_{2}$
$[C]_{so l} = [C]_{urea}$
$\frac{5.25}{M} = \frac{1.5}{60}$
$\Rightarrow M = \frac{5.25 \times 60}{1.5}$
$= 210$

Q. A 5.25% solution of a substance is isotonic with 1.5% solution of urea (molar mass =60gmol−1 ) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0gcm−3 , molar mass of the substance will be :

90.0gmol−1

11.50gmol−1

105.0gmol−1

210.0gmol−1