Test Index

JAMB Chemistry 2010 Paper

© africaexams.com

Question : 18

Total: 49

Calculate the volume of 0.5 mol dm-3 H2So4 that is neutralized by 25 cm3 of 0.1 mol dm-3 NaOH?

5.0 cm³
2.5 cm³
0.4 cm³
0.1 cm³

Solution:

Concept:

The neutralization reaction between

\mathrm{H}_2\mathrm{SO}_4

and

\mathrm{NaOH}

has a mole ratio of

1:2

.

Explanation:

Moles of

\mathrm{NaOH} = 0.1 \, \mathrm{mol\,dm^{-3}} \times 0.025 \, \mathrm{dm^3} = 0.0025 \, \mathrm{mol}

.

From

\mathrm{H}_2\mathrm{SO}_4 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{H}_2\mathrm{O}

, moles of

\mathrm{H}_2\mathrm{SO}_4

neutralized

= \frac{0.0025}{2} = 0.00125 \, \mathrm{mol}

.

Volume of

\mathrm{H}_2\mathrm{SO}_4 = \frac{0.00125}{0.5} = 0.0025 \, \mathrm{dm^3} = 2.5 \, \mathrm{cm^3}

.

Answer:

2.5 \, \mathrm{cm^3}

, which is option B.

© africaexams.com

Go to Question:

Prev Question

Next Question