[18" g glucose "(C_(6)H_(12)O_(6))" dissolved in "500" g of water.Find depression in freezing "],[" point "(Delta Tf).]

1.8g of fructose (C_(6)H_(12)O_(6)) is added to 2 kg of water. The freezing point of the solution is

The molal freezing point constant of water is 1.86 ^(@) C m^(-1) . If 342 g of cane sugar (C_(12)H_(22)O_(11)) is dissolved in 1000 g of water, the solution will freeze at

The molal freezing point constant of water is 1.86 K m^(-1) . If 342 g of cane sugar (C_(12)H_(22)O_(11)) is dissolved in 1000 g of water, the solution will freeze at

The molal freezing point constant of water is 1.86 K m^(-1) . If 342 g of cane sugar (C_(12)H_(22)O_(11)) is dissolved in 1000 g of water, the solution will freeze at

The molal freezing point constant of water is " 1.86 Km"^(-1) . If 342 g of cane sugar (C_(12)H_(22)O_(11)) is dissolved in 1000 g of water, the solution will freeze at

Find the elevation in boiling point and (ii) depression in freezing point of a solution containing 0.520g glucose (C_(6)H_(12)O_(6)) dissolved in 80.2g of water. For water K_(f)= 1.86k//m. K_(b)=0.52 K//m.