Question

Find the freezing point of solution containing 3.6 g of glucose dissolved in 50g ofwater kg of water

Answer 1

Answer:

The molecular weight of glucose C

6

H

12

O

6

=6(12)+12(1)+6(16)=180 g/mol.

The number of moles of glucose =

180g/mol

0.625g

=0.00347moles

Mass of water =102.8g=0.1028kg

The molality of glucose m=

0.1028kg

0.00347moles

=0.0338mol/kg

The depression in the freezing point ΔT

f

=K

f

m=1.87Kkg/mol×0.0338mol/kg=0.0632K

Freezing point of water =273 K

The freezing point of the solution =273+0.0632=273.0632K