Question

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL⁻¹, then what shall be the molarity of the solution?

Answer 1

Molar mass of ethylene glycol [C₂H₄(OH)₂] = 2 × 12 + 6 × 1 + 2 × 16

= 62 g mol⁻¹

Number of moles of ethylene glycol = `(222.6  "g")/(62  "g mol"^(-1))`

= 3.59 mol

Therefore, molality of the solution = `(3.59  "mol")/(0.200  "kg")`

= 17.95 m

Total mass of the solution = (222.6 + 200) g

= 422.6 g

Given,

Density of the solution = 1.072 g mL⁻¹

∴ Volume of the solution = `(422.6  "g")/(1.072  "g mL"^(-1))`

= 394.2 mL

= 0.3942 L

⇒ Molarity of the solution = `(3.59  "mol")/(0.3942  "L")`

= 9.11 M

Answer 2

Molality (m) of the solution is given by

`m = w/(M') xx 1000/(w')`

In the present case, w = 222.6 g, M' = 62 g mol⁻¹, w' = 200 g (Mol. mass of glycol = 62)

∴ `m = (w xx 1000)/(M' xx w')`

= `(222.6 xx 1000)/(62 xx 200)`

= 17.95 mol kg⁻¹

`"Volume of solution" = "Mass"/"Density"`

= `(222.6 + 200)/(1.072)`

= 394.2 mL

Molarity is given by

`w = (M xx M' xx v)/1000`

∴ `M = (w xx 1000)/(M' xx v)`

= `(222.6 xx 1000)/(62 xx 394.2)`

= 9.11 mol L⁻¹