Question
CBSEENCH12010696

The density of a solution prepared by dissolving 120 g of urea (mol. Mass = 60 u ) in 1000g of water is 1.15 g/mL. The molarity of this solution is

  • 0.50 M

  • 1.78 M

  • 1.02 M

  • 2.05

Solution

D.

2.05

molarity space equals space fraction numerator moles space of space space solute over denominator volume space of space solution end fraction
Total mass of solution
= 1000 g water +120 g urea
 = 1120 g
Density space of space solution space space equals space 1.15 space straight g divided by ml
Thus comma space volume space of space solution space equals space mass over density space equals space fraction numerator 1120 space straight g over denominator 1.15 space straight g divided by mL end fraction
space equals space 973.91 space mL space equals space 0.974 space straight L
Molarity space equals space fraction numerator 2 over denominator 0.974 end fraction space equals space 2.05 space mol space straight L to the power of negative 1 end exponent
space equals space 20.5 space straight M

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