CBSE Class 12 · 2007 · Delhi · SolutionsShort Answer

An antifreeze solution is prepared from 222.6 g of ethylene glycol (C

_{2}

_{4}

(OH)

_{2}

) and 200 g of water. Calculate the molality of the solution. If the density of this solution is 1.072 g mL

^{- 1}

, what will be the molarity of the solution?

Answer: m ≈ 17.95 mol kg⁻¹; M ≈ 9.11 mol L⁻¹

Solution

Moles of glycol (

M = 62

n = \frac{222.6}{62} = 3.59 mol

Molality:

m = \frac{3.59}{0.200}

m = 17.95 mol kg^{- 1}

Total mass of solution =

222.6 + 200 = 422.6

g; volume:

V = \frac{422.6}{1.072} = 394.2 mL = 0.3942 L

Molarity:

M = \frac{3.59}{0.3942}

M = 9.11 mol L^{- 1}

CBSE Class 12 Board · 2007 · Delhi · Subjective