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Question
CBSEENCH12011232
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A 0.1  molal aqueous solution  of a weak  acid is 30%  ionised. if Kf for water is 1.86oC/m, the freezing point of the solution will be 

  • -18oC

  • -0.54oC

  • -0.36oC

  • -0.24oC

Solution

D.

-0.24oC

HA space rightwards arrow space space space space space space space space space space straight H to the power of plus space plus space straight A to the power of minus 1 minus straight alpha space space space space space space space space space space space space space space space straight alpha space space space space space space space space straight alpha 1 minus 0.3 space space space space space 0.3 space space space space space space space space 0.3 space straight i space equals space 1 minus 0.3 space plus 0.3 space plus 0.3 straight i space equals space 1.3 therefore comma increment straight T subscript straight f space equals space 1.3 space straight x space 1.86 space straight x space 0.1 space equals 0.2418 to the power of straight o straight C straight T subscript straight f space equals space 0 minus 0.2418 to the power of straight o straight C

Question
CBSEENCH12011032
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A 1000C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be

  • 1000C

  • 1020C

  • 1030C

  • 1010C

Solution

D.

1010C

According space to space Raoult apostrophe straight s space law space of space paritial space pressure. fraction numerator straight p subscript straight A superscript 0 minus straight p subscript straight s over denominator straight p subscript straight s end fraction space equals straight n subscript straight B over straight n subscript straight A rightwards double arrow space fraction numerator 760 minus 732 over denominator 732 end fraction space equals space fraction numerator straight W subscript straight B space straight x space straight M subscript straight A over denominator straight M subscript straight B space xW subscript straight A end fraction rightwards double arrow space 28 over 732 space equals space fraction numerator 6.5 space straight x space 18 over denominator straight M subscript straight B space straight x space 100 end fraction rightwards double arrow straight M subscript straight B space equals space 30.6 therefore space increment straight T subscript straight b space equals 0.52 space straight x fraction numerator 6.5 space straight x 1000 over denominator 30.6 space space straight x space 100 end fraction space equals 1.10 therefore space Boiling space point space equals 100 plus 1.10 equals 101.1 degree straight C space almost equal to space 101 degree straight C

Question
CBSEENCH12011070
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A mixture of gases contains H2 and O2 gases in the ratio of 1:4 (w/w).what is the molar ratio of the two gases in the mixture?

  • 1:4

  • 4:1

  • 16:1

  • 2:1

Solution

B.

4:1

Let the mass of H2 gas be x g and mass of O2 gas 4x g
Molar      H2 :    O2
mass      2   :     32

i.e          1:      16
therefore, Molar ratio =  straight n subscript straight H subscript 12 end subscript over straight n subscript straight O subscript 2 end subscript space equals space fraction numerator x divided by 2 over denominator 4 x divided by 32 end fraction space equals space fraction numerator left parenthesis x space right parenthesis space x space 32 over denominator space 2 space x space 4 x end fraction space equals space 4 over 1 space equals space 4 colon 1

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Question
CBSEENCH12011386
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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

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Question
CBSEENCH12011404
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A solution of acetone in ethanol:

  • shows a negative deviation from Raoult's law

  • shows a positive deviation from Raoult's law

  • behave likea near ideal solution

  • obeys Raoult's law

Solution

B.

shows a positive deviation from Raoult's law

A solution ofacetone in ethanol shows a positive deviation from Raoult's law due to miscibility of these two liquids with a difference of polarity and length of the hydrocarbon chain.

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