200 ~mL of an aqueous solution of a protein contains its 1.26 ~g . The osmotic pressure of this solution at 300 ~K is found to be 2.57 × 10^ -3 bar. The molar mass of protein will be

Correct answer: D. 61038 ~g ~mol^ -1 aligned & π=C R T &= w × 1000 M × V( in mL ) T & [ = n × 1000 V( in mL ) and n=(w)/(M)] &= 1.26 × 1000 × 0.083 × 300 2.57 × 10^ -3 × 200 &=61038 ~g ~mol^ -1 aligned

NEET 2011

Exam: NEET
Year: 2011
Subject: Chemistry
Chapter: Solutions

NEET 2011 · Chemistry · Solutions

200 ~mL of an aqueous solution of a protein contains its 1.26 ~g ....

200 \mathrm{~mL}

of an aqueous solution of a protein contains its

1.26 \mathrm{~g}

. The osmotic pressure of this solution at

300 \mathrm{~K}

is found to be

2.57 \times 10^{-3}

bar. The molar mass of protein will be

Options

A $51022 \mathrm{~g} \mathrm{~mol}^{-1}$
B $122044 \mathrm{~g} \mathrm{~mol}^{-1}$
C $31011 \mathrm{~g} \mathrm{~mol}^{-1}$
D $61038 \mathrm{~g} \mathrm{~mol}^{-1}$

Correct answer

D. $61038 \mathrm{~g} \mathrm{~mol}^{-1}$

Step-by-step solution

\begin{aligned}& \because \pi=C R T \\ &=\frac{w \times 1000}{M \times V(\text{\text{ in } mL })} \timesR T \\ & \quad\left[\becauseC=\frac{n \times 1000}{V(\text{\text{ in } mL })} \text{ and }n=\frac{w}{M}\right]\\ &=\frac{1.26 \times 1000 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 200}\\ &=61038 \mathrm{~g} \mathrm{~mol}^{-1} \end{aligned}

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