A 5 .25 % solution of a substance is isotonic with a 1.5 % solution of urea (molar mass = 60 g mol-1)in the same solvent. If the densities of both the solution are assumed to be equal to 1.0 g cm-3 molar mass of the substance will be

Question

A 5 .25 % solution of a substance is isotonic with a 1.5 % solution of urea (molar mass = 60 g mol-1)in the same solvent. If the densities of both the solution are assumed to be equal to 1.0 g cm-3 molar mass of the substance will be (A) 210.0 g mol-1 (B) 90.0 g mol-1 (C) 115.0 g mol-1 (D) 105.0 g mol-1

Tardigrade Admin · Accepted Answer

Isotonic solutions have the same molar concentration. Hence,
Molar cone. of the substance=Molar cone of urea

\frac{52.5 g L ^{- 1}}{M} = \frac{15 g L ^{- 1}}{60 g m o l ^{- 1}}

(100 g solution = 100 mL as d = 1 g

m o l^{- 1}

)
M =

\frac{52.5 \times 60}{15}

= 210 g mol

^{- 1}

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

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

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

$115.0 g m o l^{- 1}$

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

Isotonic solutions have the same molar concentration. Hence,
Molar cone. of the substance=Molar cone of urea
$\frac{52.5 g L ^{- 1}}{M} = \frac{15 g L ^{- 1}}{60 g m o l ^{- 1}}$
(100 g solution = 100 mL as d = 1 g $m o l^{- 1}$ )
M = $\frac{52.5 \times 60}{15}$ = 210 g mol $^{- 1}$

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

210.0gmol−1

90.0gmol−1

115.0gmol−1

105.0gmol−1