Q. 84.4( 42 Votes )

# An antifreeze solution is prepared from 222.6 g of ethylene glycol (C_{2}H_{6}O_{2}) 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 :

Given:

Mass of ethylene glycol (C_{2}H_{6}O_{2}) = 222.6 g

Mass of water = 200 g

Density, d = 1.072 g/ml

To find: Molality and Molarity of solution

Formula:

Molality

Molarity, M_{o}

Density, d

Solution:

Calculation of Molality:

⇒ Molecular Mass of ethylene glycol (C_{2}H_{6}O_{2}) = [12 × 2] + [6 × 1] + [16 × 2]

= 24 + 6 + 32

= 62 g

⇒ Number of moles of ethylene glycol (C_{2}H_{6}O_{2}) = [222.6/62]

= 3.59 moles

⇒ Mass of water = 200 g

⇒ Molality

= 17.95 m

Calculation of Molarity:

⇒ Total mass of solution = [222.6 + 200]

= 422.6 g

From density formula we can find out the volume required.

⇒ Volume of solution, V = [422.6 /1.072]

= 394.216 ml

⇒ Molarity, M_{o}

⇒ Molarity, M_{o}

⇒ Molarity = 9.1067

⇒ Molarity ≈ 9.11 M

Therefore the Molality and Molarity of the solution is as follows:

Molality = 17.95 m

Molarity = 9.11 M

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Match the terms given in Column I with expressions given in Column II.