Question

Mass of beaker:13.50g

Mass of unknown: 4.15g

Mass of water: 15.7g

Freezing point of pure water: 0.15 C

Freezing point of unknown solution: -4.35 C

Assume you use the ice bath as described in the experiment to obtain the freezing point of pure water as well as the freezing point of your sample solution above, and measure the final freezing point temperatures indicated below

a.) What is the Freezing Point Depression for this experiment (how much did the freezing point decrease)?

b.)Based on this freezing point depression, what is the Molality of the Solution?

c.)Based on that molality what is the Moles of Unknown (you do know the kg of water, right)?

d.)What then is the Molecular Weight of the Unknown?
(you can look at units you need to figure this out)

Answer 1

Given mass of the beaker= 13.50 g, mass of water with beaker = 15.7 g mass of water= 2.20 g, mass of unknown= 4.15 g, freezing point of pure water T_o=0.15 ^oC, freezing point of unknown solution T_s= -4.35^oC Depression in freezing point= T_o-T_s= 0.15-(-4.35)= 4.50 ^oC, depression in freezing point= K_f* molality, Molality= 4.50/1.86 mol/ Kg​ Molality= 2.42 mol/Kg, molality= no. of moles / Kg of water, No. of moles = 2.42 , we know that No of moles= weight of the unknown solution/ molecular weight, m.wt= weight/ moles= 4.35/2.42= 1.797 g, molecular weight of unknown solution = 1.797 g