Question

A solution prepared by dissolving 513 mg of a sugar (a molecular compound and a nonelectrolyte) in 1.00 g of water froze at -2.79 °C. What is the molar mass of this sugar? The value of Kf is 1.86°C/m. Answer in g/mol.

Answer 1

DTf = i*kf*m

DTf = T0-Tf = 0 - - 2.79 = 2.79 c

DTf = i*Kf*m

i = vanthoff factor of sugar = 1

Kf = frreezing point constant of water = 1.86 K/m

molality(m) = (w/mwt)*(1000/W)

             = (0.513/x)*(1000/1)

2.79 = 1*1.86*(0.513/x)*(1000/1)

molarmass of sugar = x = 342 g/mol