Test Index

WAEC Chemistry 2025 Paper

© africaexams.com

Question : 22

Total: 50

The number of moles of KOH present in 200cm

^3

^{-3}

solution is [KOH = 56 g mol

^{-1}

]

2.01
1.12
0.02
0.01

Solution:

Concept:

Convert the concentration and volume into mass, then divide by the molar mass.

Explanation:

Concentration of

\mathrm{KOH} = 5.6\ \mathrm{g\,dm^{-3}}

Volume

= 200\ \mathrm{cm^3} = \frac{200}{1000}\ \mathrm{dm^3} = 0.2\ \mathrm{dm^3}

Mass of

\mathrm{KOH} = 5.6 \times 0.2 = 1.12\ \mathrm{g}

Moles of

\mathrm{KOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{1.12}{56} = 0.02\ \mathrm{mol}

Answer:

0.02

mol, which is option C.

© africaexams.com

Go to Question:

Prev Question

Next Question