Question

Boiling point of water at 750 mm \[\ce{Hg}\] pressure is 99.68°C. How much sucrose (Molar mass = 342 g mol⁻¹) is to be added to 500 g of water such that it boils at 100°C? (K_b for water = 0.52 K kg mol⁻¹).

Answer 1

Given,

Boiling point of pure water = 99.68°C

Elevation in boiling point, (ΔT_b) = 100 − 99.68 = 0.32°C

K_b = 0.52 K kg mol⁻¹ 

Molar mass of sucrose= 342 g mol⁻¹

Amount of water (W_A) = 500 g

We know that ΔT_b = K_b × molality   ...(i)

`"Molality" = ("Mass of solute" xx 1000)/("Molar mass of solute" xx "Mass of solvent")("g")`

Let, x g of sucrose be added

∴ Molalit = `(x xx 1000)/(342 xx 500)`

= `x/171`

Putting the value in Eq. (i), we get

`0.32 = 0.52 xx x/171`

`x = (0.32 xx 171)/(0.52)`

= 105.23 g

Therefore, 105.23 g of sucrose must be added.