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=0.4g,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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