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Solutions question

2010 · Q103
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Solutions question

2010 · Q103

NEETChemistrySolutionsMCQ+4 / −1
A solution of sucrose (molar mass = 342 g mol−-−) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol−-−1)
  1. A
    −-− 0.372oC
  2. B
    −-− 0.520oC
  3. C
    + 0.372oC
  4. D
    −-− 0.570oC
View written solutionFree

Correct answer: A

Depression in freezing point,

Δ\Delta ΔTf = Kf ×\times× m

m = wBMB×1000WA{{{w_B}} \over {{M_B}}} \times {{1000} \over {{W_A}}}MB​wB​​×WA​1000​

= 68.5×1000342×1000{{68.5 \times 1000} \over {342 \times 1000}}342×100068.5×1000​

Δ\Delta ΔTf = 1.86 ×\times× 68.5342{{68.5} \over {342}}34268.5​

= 0.372 oC

∴\therefore∴ Tf = 0 - 0.372 oC = - 0.372 oC

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