1.00 g of non-electrolyte solute (molar mass 250 g mol^-1 ) was dissolved in 51.2 g of benzene. If the freezing point depression constant, K_f of benz - Chemistry MCQ Chemistry | ExamTest.live

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1.00 g of non-electrolyte solute (molar mass 250 g mol^{- 1}) was dissolved in 51.2 g of benzene. If the freezing point depression constant, K_{f} of benzene is 5.12 K kg mol^{- 1}, the freezing point of benzene will be lowered by :

A $0.4 K$
B $0.3 K$
C $0.5 K$
D $0.2 K$

Solution & Step-by-step Explanation

The depression in freezing point is Δ T_{f} = K_{f} \times m (molality). m = \frac{Mass of solute}{Molar mass} \times \frac{1000}{Mass of solvent (g)} m = \frac{1.00}{250} \times \frac{1000}{51.2} = \frac{4}{51.2} mol/kg . Δ T_{f} = 5.12 \times \frac{4}{51.2} = 5.12 \times 0.078125 = 0.4 K .

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1.00 g of non-electrolyte solute (molar mass 250 g mol^{- 1}) was dissolved in 51.2 g of benzene. If the freezing point depression constant, K_{f} of benzene is 5.12 K kg mol^{- 1}, the freezing point of benzene will be lowered by :

0.4 K

0.3 K

0.5 K

0.2 K

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$1.00 g$ of non-electrolyte solute (molar mass $250 g mol^{- 1}$ ) was dissolved in $51.2 g$ of benzene. If the freezing point depression constant, $K_{f}$ of benzene is $5.12 K kg mol^{- 1}$ , the freezing point of benzene will be lowered by :

Solution & Step-by-step Explanation

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1.00g of non-electrolyte solute (molar mass 250g mol−1) was dissolved in 51.2g of benzene. If the freezing point depression constant, Kf​ of benzene is 5.12K kg mol−1, the freezing point of benzene will be lowered by :

Solution & Step-by-step Explanation

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Discussion

$1.00 g$ of non-electrolyte solute (molar mass $250 g mol^{- 1}$ ) was dissolved in $51.2 g$ of benzene. If the freezing point depression constant, $K_{f}$ of benzene is $5.12 K kg mol^{- 1}$ , the freezing point of benzene will be lowered by :