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 solutions are assumed to be equal to 1.0 ~g ~cm^ -3 , molar

Correct answer: D. 210.0 ~g ~mol^ -1 Solutions with the same osmotic pressure are isotonic Let the molar mass of the substance be M π_1=C_1 RT=C _2 RT=π_2 So, C_1=C_2 As density of the solutions are same So (5.25)/(M)=(15)/(60) M=(5.25 × 60)/(1.5)=210 Hence (D) is correct

JEE Main 2007

Exam: JEE Main
Year: 2007
Subject: Chemistry
Chapter: States of Matter

JEE Main 2007 · Chemistry · States of Matter

A 5.25 \% solution of a substance is isotonic with a 1.5 \%...

5.25 \%

solution of a substance is isotonic with a

1.5 \%

solution of urea \left(\right.

molar mass

=60 \mathrm{~g} \mathrm{~mol}^{-1}

) in the same solvent. If the densities of both the solutions are assumed to be equal to

1.0 \mathrm{~g} \mathrm{~cm}^{-3}, molar mass of the substance will be

Options

A $90.0 \mathrm{~g} \mathrm{~mol}^{-1}$
B $115.0 \mathrm{~g} \mathrm{~mol}^{-1}$
C $105.0 \mathrm{~g} \mathrm{~mol}^{-1}$
D $210.0 \mathrm{~g} \mathrm{~mol}^{-1}$

Correct answer

D. $210.0 \mathrm{~g} \mathrm{~mol}^{-1}$

Step-by-step solution

Solutions with the same osmotic pressure are isotonic Let the molar mass of the substance be

M

\pi_1=\mathrm{C}_1

\mathrm{RT}=\mathrm{C}

_2 \mathrm{RT}=\pi_2

So,

\mathrm{C}_1=\mathrm{C}_2

As density of the solutions are same So

\frac{5.25}{M}=\frac{15}{60}

M=\frac{5.25 \times 60}{1.5}=210

Hence (D) is correct

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