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:-A. 0.4 KB. 0.3 KC. 0.5 KD. 0.2 K

1.00 g of a non-electrolyte solute (molar mass 250g `mol^(–1)`) was

1.	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:-A. 0.4 KB. 0.3 KC. 0.5 KD. 0.2 K
Answer» Correct Answer - A `DeltaT_(f)=molalityxxK_(f)` `DeltaT_(f)=((1xx1000)/(250xx51.2))xx5.12` `DeltaT_(f)=0.4K`

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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:-A. 0.4 KB. 0.3 KC. 0.5 KD. 0.2 K

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