1.00 g of a non-electrolyte solute (molar mass 250g 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

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0.3 K

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0.5 K

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0.2 K

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Solution

The correct option is A 0.4 K
$Δ T_{f} = K_{f} \times \frac{W_{1} \times 1000}{W_{2} \times M_{1}} w h e r e W_{1} = W e i g h t o f t h e s o l u t e W_{2} = W e i g h t o f s o l v e n t M_{1} = M o l a r m a s s o f s o l u t e K_{f} = F r e e z i n g p o i n t d e p r e s s i o n c o n s t a n t N o w, Δ T_{f} = 5.12 \times \frac{1 \times 1000}{51.2 \times 250} Δ T_{f} = 0.4 K$

Hence, the correct option is $A$

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