Question

A 0.194 g sample of a nonvolatile solid solute dissolves in 9.82g of cyclohexane.The change in the freezing point of the solution is 2.94 Celsius.

a). What is the molarity of the solute in the solution?

b). Calculate the molar mass of the solute to the correct number of sig fig.

c). The same mass of solute is dissolved in 9.82g of t-butanol instead of cyclohexane. What is the expected freezing point change of this solution?

Answer 1

a) dT = i x Kb x m is formula ,

2.94 = 1 x 20.2 x m                ( i = 1 for non dissociative solids, Kb = 2.94 )

molality = 0.14555 = Moles of solute / solvent mass in Kg

0.14555= moles of solute / 0.00982

moles of solute = 0.001429

to get Molarity we need density of solution , if we assume 1g/ml then

solution mass = 9.82+0.194 = 10.014 kg= 0.01 Kg , solution volume = 0.01 L

Molarity = ( moles of solute ) / solution vol in L = 0.001429 /0.01 = 0.1429 M

b) Molar mass = mass /Moles = 0.194/0.001429 = 135.76 g/mol

c) for t-butanol Kf = 9.1

hence dT = 1 x 9.1 x 0.14555 = 1.324

Freezing point of solution = cyclohexane Freezing point - 1.324 = 6.47 -1.324 = 5.15 C