Question

1. ) An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the elements is
10.8 g cm-. Calculate the number of atoms in 108 g of the element.
To = 273²
2. ) A 4% solution (w/w) of sucrose (M = 342 g mol-l) in water has a freezing point of 27115 K. Calculate
the freezing point of 5% glucose (M = 180 g mol-') in water. (Given : Freezing point of pure water =
273.15 K).
​

Answer 1

Explanation:

1) Given- fcc lattice, density = 10.8 g cm-3 = β, edge length = 300pm = 300 × 10-12 m = 300 x 10-10cm = a

Find- no. Of atoms in 108g of element

Using the formula for density,

Being fcc lattice Z = 4

Putting all the values on equation we get, 10.8 =

M= 175.6 g

In 175.6 g, 6.022 X 1023 atoms are present

Therefore, in 108g, no. Of atoms will be

= 0.61 x 1023 atoms

2) 4% solution (w/w) of sucrose means 4g of sucrose in 96g water

Wsolvent = 96g, wsolute= 4g, Msolute= 342 g mol-1

Using the formula, ∆Tf = kf x m

(273.15 – 271.15) =

Kf = 16.39 K

For 5% solution (w/w) of sucrose means 5g of sucrose in 95g water

Wsolvent = 95g, wsolute= 5g, Msolute= 180 g mol-1

Using the formula, ∆Tf = kf x m

∆Tf =

K

= 0.48 K

Tf = Tf0 - ∆Tf = (273.15 – 0.48) K = 272.67 K