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Solutions

Question
CBSEENCH12011386

A solution containing 10 g per dm3 is urea (molecular mass = 60 g mol-1) is isotonic with a molecular mass of this non-volatile solute. The molecular mass of this of this non-volatile solute is:

  • 250 g mol-1

  • 300 g mol-1

  • 350 g mol-1

  • 200 g mol-1

Solution

B.

300 g mol-1

10 g per dm3 of urea is isotonic with 5% solution of a non-volatile solute. Hence, between this solution osmosis is not possible so, their molar concentrations are equal to each other,
Thus, molar concentration of urea solution
space equals space fraction numerator 10 straight g divided by dm cubed over denominator Mol. space wt. space of space urea end fraction
space equals 10 over 60 straight M space equals space 1 over 6 straight M
Molar space concentration space of space 5 percent sign space non minus volatile space solute

equals space fraction numerator 50 space straight g divided by dm cubed over denominator mol. wt. space of space non minus volatile space solute end fraction
space equals space 50 over straight m space straight M
Both space solution space are space isotonic space to space each space other comma space therefore comma space
1 over 6 space equals space 5 over straight m
straight m space equals space 50 space straight x space 6 space equals space 300 space straight g space mol to the power of negative 1 end exponent