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NEET levelSingle correct optionDifficulty 1 of 3so059

Glucose (molar mass $180 g mo l^{- 1}$ ), $1.8 g$ , is dissolved in $100 g$ of water. Taking $K_{f} = 1.86 K kg mo l^{- 1}$ , the depression in freezing point is

A $0.186 K$
B $1.86 K$
C $0.372 K$
D $0.0186 K$

Answer: A

Moles of glucose

= 1.8/180 = 0.010

; molality

= 0.010/0.100 = 0.10 mol k g^{- 1}

.

Δ T_{f} = 1.86 \times 0.10 = 0.186 K

.

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