At T (K ) g of a non-volatile solid (molar mass 78 g mol^-1 ) when added to 0.5 kg water, lowered its freezing point by 1.0^ C. What is x (in (g ) ? (K_f ) of water at T (K )= (1.86 KKg mol^-1 )

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Correct answer: B. (b ) Molality of the solution= ( x78 ) 1000500= x39 ( m ) T_f=m (K_f )_H_2O ∴ x39= 11.86 ∴x= 391.86=20.96

$At T (K) \times g of a non-volatile solid (molar mass 78 g mol^{- 1}) when added to 0.5 kg water, lowered its freezing point by 1.0^{\circ} C . What is x (in (g) ? (K_{f}) of water at T (K) = (1.86 KKg mol^{- 1})$

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