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WAEC Chemistry 2013 Paper

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Question : 20

Total: 47

The volume of 0.25 moldm-3 solution of KOH that would yield 6.5g of of solid KOH on evaporation is (K = 39.0; o = 16.0; H = 1.00)

464.30 cm³
625.00 cm³
1000.00 cm³
2153.80 cm³

Solution:

Concept:

The amount of solute in moles remains the same before and after evaporation, so moles = mass ÷ molar mass = concentration × volume (in

\mathrm{dm^3}

).

Explanation:

Molar mass of KOH =

39.0 + 16.0 + 1.00 = 56.0\ \mathrm{g\,mol^{-1}}

Moles of KOH in

6.5\ \mathrm{g}

is

n = \frac{6.5}{56.0} = 0.11607\ \mathrm{mol}

Using

C = \frac{n}{V}

, rearrange to

V = \frac{n}{C}

V = \frac{0.11607}{0.25} = 0.46428\ \mathrm{dm^3}

Convert to

\mathrm{cm^3}

:

0.46428 \times 1000 = 464.28\ \mathrm{cm^3} \approx 464.30\ \mathrm{cm^3}

Answer:

A.

464.30\ \mathrm{cm^3}

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