The molal freezing point constant of water is $1.86 K m^{- 1}$ . If 342 g of cane sugar $C_{12} H_{22} O_{11}$ is dissolved in 1000 g of water, the solution will freeze at:

- {1.86}^{o} C

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{1.86}^{o} C

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- {3.92}^{o} C

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{2.42}^{o} C

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Solution

The correct option is C $- {1.86}^{o} C$
$Δ T_{f} = \frac{1000 \times K_{f} \times w}{M w \times W}$
$= \frac{1000 \times 1.86 \times 342}{342 \times 1000} = 1.86 K$

$∴ T_{f} = 0 - 1.86 = - {1.86}^{o} C$

Hence, the correct option is $A$

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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:

The correct option is C −1.86oCΔTf=1000×Kf×wMw×W =1000×1.86×342342×1000=1.86 K∴Tf=0−1.86=−1.86oCHence, the correct option is A

The molal freezing point constant of water is $1.86 K m^{- 1}$ . If 342 g of cane sugar $C_{12} H_{22} O_{11}$ is dissolved in 1000 g of water, the solution will freeze at:

The correct option is C $- {1.86}^{o} C$
$Δ T_{f} = \frac{1000 \times K_{f} \times w}{M w \times W}$
$= \frac{1000 \times 1.86 \times 342}{342 \times 1000} = 1.86 K$

$∴ T_{f} = 0 - 1.86 = - {1.86}^{o} C$

Hence, the correct option is $A$