Question

there two questions :)

Enter your answer in the provided box. What is the molality of a solution containing 4.37g of urea [(NH2)2 CO] in 176.5 g of water?

Answer 1

Molality - moles ot solute - kg of solvent Hene, molecular weight or wrea = (14+1X2)X2+12+ 16 g/mol = 60 g/mol : Moles of wrea present - 4.37 % -7.28 x162 mol 60 g/mol Amount of solvent = 176.5 g = 176.5 x163 kg (5:19=162 kg) = 0.1765 kg . Molality - 7.28x10 m = 10.612m Malaya 0.1765m 10.412 m (ANSWER)

Das The formula for freezing point depression is — -4T -- kg x m where, 4T = freezing point depnession Ky = molal freezing point depression of solvent M = molality of the solution Given, OT = 15.300+10b (5.166-5.500) °C = -0.334 oe Kg = 5.12 °C/m i -0.334 = – 5.12 x m + m = 6.523 x102 So, 1000g benzene contains 6.523 x 152 mol solute 100g 1 6.523 x102x100 1000 = 6.523 x103 : 6.523 x1Ở mol = 0.7909 -3 - 1 mol 0.790 x1 = 6.523x10-3 g = 12lolla :. Molality - 16,53x162 m Molan may = 121.11 g/mol