The molal freezing point constant of water is $$1.86$$ K $$m-1$$. If $$342$$ g of cane sugar (C12H22O11) is dissolved in $$1000$$ g of water, the solution will freeze at – $$X0$$ C, the value of X is

∆$$Tf=1000$$×Kf×WMw $$=1000$$×$$1.86$$×$$342100$$×$$342=1.86$$ ∴ $$Tf=0-1.86=-1.86$$° C

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Question

The molal freezing point constant of water is 1.86 K m^-1. If 342 g of cane sugar (C₁₂H₂₂O₁₁) is dissolved in 1000 g of water, the solution will freeze at – X⁰ C, the value of X is

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Solution

$∆ T_{f} = \frac{1000 \times K_{f} \times W}{M w} = \frac{1000 \times 1.86 \times 342}{100 \times 342} = 1.86 ∴ T_{f} = 0 - 1.86 = - 1.86 ° C$

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