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WAEC Chemistry 2013 Paper

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

Validate

Solution:

number of moles = molar concentration x volume( $\mathrm{dm}^3$ )

molar concentration(C) = 0.25mol/dm3, volume = ? and number of mole(n) = $\; \frac{\text{mass}}{\text{molar mass}}$ = $\; \frac{6.5}{56}$ ( where 56g/mol = molar mass of KOH)

$\; \frac{6.5}{56}$ = 0.25 x V

V = $\; \frac{0.11607}{0.25}$ = 0.46428dm^3 → 464.3cm^3

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