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NEET UG CHEMISTRY MCQ: SOLUTIONS

1 Hard

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Application +4 | -1

A solution of protein (extracted from crabs) was prepared by dissolving 0.75 g in

125\;\mathrm{c}{\mathrm{m}^3}

of an aqueous solution. At

4^\circ C

an osmotic pressure rise of 2.6 mm of the solution was observed. Then molecular weight of protein is : (Assume density of solution is

1.00\;g/\mathrm{c}{\mathrm{m}^3}

)

View answer and explanation

Correct option: B

5.4 \times {10^5}

Option B (

5.4 \times {10^5}

) is correct because the molecular weight of the protein is calculated using the osmotic pressure formula.

More Questions from SOLUTIONS

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Consider the solution of ethanol. In it Assertion: Isotonic solutions don't show net phenomenon of osmosis.

Reason: Isotonic solutions have equal osmotic pressure at constant temperature. 0.2 molal aqueous solution of a monobasic acid is 60% ionised. If Kb for water is

0.52\,K\,kg\,mo{l^{ - 1}}

then the boiling point of the solution will be The vapour pressures of pure liquids

A

B

are 400 and 600 mmHg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid

B

is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components

A

B

in vapour phase, respectively are What is molecular weight of a substance whose 7.0% by mass solution in water freezes at

--{0.93^0}C

{K_f}

1.86\mathrm{ }K\mathrm{ }kg\mathrm{ }mo{l^{ - 1}}

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. the vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be