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Home  >>  AIMS  >>  Class12  >>  Chemistry  >>  Liquid Solutions
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The molal freezing point constant for water is 1.86 K.molality–1. If 34.2 g of cane sugar ($C_{12}H_{22}O_{11}$) are dissolved in 100g of water, the solution will freeze at __$^{\large\circ}C$

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$W_2$ = 34.2 g
$W_1$ = 100 g
$K_f = 1.86 K.molality^{–1}$
$M_2 = 342 g mol^{−1}$
We know $\Delta T_f=K_fm=\large\frac{K_f\times W_2\times 1000}{W_1 \times M_2}$
$\Rightarrow \large\frac{1000\times 1.86\times 34.2}{100\times 342}$
$T_f=0-1.86=-1.86^{\large\circ}C$
Hence (B) is the correct answer.
answered Jun 12, 2014 by sreemathi.v
 

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