1) Calculate the freezing point of a solution that contains 5.37 g of glucose, a

Question

1) Calculate the freezing point of a solution that contains 5.37 g of glucose, a non electrolyte, dissolved in 118 of water. (Please show work)

2) If 5.33 g of Ethylene Glycol are added to 107.5 g of water what is the freezing point of the solution? (answer should be negative)

3) If 34.1 g of benzene, a nonelectrolyte, are added 929 g of water. What is the freezing point of the solution inn Celsius? Answer should be negative

4) The molality of a sample, a nonelectrolyte, is 0.01547 mol/kg and the mass of the solute is 2.15 g and the mass of the solvent is 952 g. What is the molar mass of the sample g/ mol ?

Explanation / Answer

1)

molar mass of glucose = 180 g/mol

mass of glucose = 5.37 g

number of mol of glucose = mass / molar mass

= 5.37 / 180

= 0.0298 mol

mass of solvent = 118 g = 0.118 Kg

molality, m = number of mol of glucose / mass of solvent in Kg

= 0.0298 mol / 0.118 Kg

= 0.2525 molal

Kf of water = 1.86 oC/m

use:

delta Tf = Kf * m

= 1.86 * 0.2525

= 0.47 oC

This is decrease in freezing point

Actual freezing point of water = 0 oC

So,

new freezing point = -0.47 oC

I am allowed to answer only 1 question at a time

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