Mass of glucose (C_6 H_ 12 O_6) required to be dissolved to prepare one litre of its solution which is isotonic with 15 ~g ~L^ -1 solution of urea (NH_3 CONH_2) is (Given: Molar mass in g mol ^ -1 C: 12, H: 1, O: 16

Correct answer: D. 45 g For isotonic solutions [osmotic pressure must be equal] π_1=π_2 C_1 RT=C _2 RT C_1=C_2 m 180 × 1 =(15)/(60 × 1) (~m is the mass of glucose) m=(180)/(4)=45 ~g

NEET 2024

Exam: NEET
Year: 2024
Subject: Chemistry
Chapter: Solutions

NEET 2024 · Chemistry · Solutions

Mass of glucose (C_6 H_ 12 O_6) required to be dissolved to prepare...

Mass of glucose

\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)

required to be dissolved to prepare one litre of its solution which is isotonic with

15 \mathrm{~g} \mathrm{~L}^{-1}

solution of urea

\left(\mathrm{NH}_3 \mathrm{CONH}_2\right)

is (Given: Molar mass in

\mathrm{g} \mathrm{mol}{ }^{-1} \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16, \mathrm{~N}: 14

)

Options

A55 g
B15 g
C30 g
D45 g

Correct answer

D. 45 g

Step-by-step solution

For isotonic solutions [osmotic pressure must be equal]

\pi_1=\pi_2

\mathrm{C}_1

\mathrm{RT}=\mathrm{C}

_2 \mathrm{RT}

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

\frac{\mathrm{m}}{180 \times 1}=\frac{15}{60 \times 1} \quad(\mathrm{~m}

is the mass of glucose)

m=\frac{180}{4}=45 \mathrm{~g}

Practice Solutions on Quantrex Academy →

More from Solutions