Q.

1.26 g of the protein is present in the aqueous solution of 200 cm3 .Calculate the molar mass of the protein,if the osmotic pressure of such solution is 2.57 x 10-3 bar at 300 K.

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61000 g mol-1

60304 g mol-1

60000 g mol-1

61009 h mol-1

answer is B.

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Detailed Solution

Given, π=2.57×10-3bar,V=200cm3=0.200 L T=300 K, R=0.082 L bar mol-1K-1 M2=w2RTπ×V M2=1.26×0.082L bar K-1mol-1×300 K2.57×10-3bar ×0.200 L =60304 g mol-1

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