Q.

A solution containing 0.5 g of a non-volatile solute in 0.2 dm3 of the solution exerts an osmotic pressure of 44.44 kPa at 300 K. Thus, molar mass of the solute is

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a

150 g mol−1

b

300 g mol−1

c

140 g mol−1

d

168 g mol−1

answer is C.

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Detailed Solution

π=wmVRTGiven, π=44.44 kPaR=8.3143 JK−1mol−1=8.3143 kPa dm3K−1mol−1w=0.4​g,m=?, V=0.2 dm3, T=300 K∴ m=wRTπV=0.5×8.3143×30044.44×0.2=140.31≈140.0 g mol−1
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