A solution consists of 200. g of a nonelectrolytic solute dissolved in 500. g of water. It freezes at -9.30°C. What is the molecular weight of the solute?

Question

Luna Chen · Accepted Answer

The Freezing point of pure water is  #0 ^0#C Depression in Freezing point is  -9.3 #0 ^0#C Step 1: Calculate the freezing point depression of the solution. #Delta#Tf = (Freezing point of pure solvent) - (Freezing point of solution)
(#0 ^0#C) - (-9.3 #0 ^0#C) = 9.3 #0 ^0#C Step 2 : Calculate the molal concentration of the solution using the freezing point depression. #Delta# Tf = (Kf) (m)  m = (9.3 #0 ^0#C) / (1.8#6^o#C /molal)
m = 5.0  molal Step 3: Calculate the molecular weight of the unknown using the molal concentration. m= 5.0 molal m = number of moles of solute /  mass of solvent (kg) number of moles of solute = ( 5.0 mol/ kg ) x 0.500 kg number of moles of solute = 2.5 moles moles of substance X  = mass  of X / Molar mass of X  2.5 moles  =  200 g / Molar Mass of X  Molar Mass of X = 200 g / 2.5 moles = 80 g/mol

Avery Adams · Answer

undefined Use the formula: $ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} $. Calculate molality. Then use the formula: $ \Delta T_f = K_f \times \text{molality} $ to find moles of solute. Finally, calculate the molecular weight using the formula: $ \text{molecular weight} = \frac{\text{mass of solute}}{\text{moles of solute}} $.