Paper 2 · chemical equilibrium

The iodine liberated in an experiment required $15.0\ cm^3$ of $0.03\ mol\,dm^{-3}$ sodium thiosulphate. The number of moles of thiosulphate used is

2.25 \times 10^{-4}

4.50 \times 10^{-4}

4.50 \times 10^{-3}

4.50 \times 10^{-2}

Explanation:

n = cV = 0.03 \times 15.0/1000 = 4.5 \times 10^{-4}\ mol

Derived from ZIMSEC Chemistry Paper 2, November 2010, Q1

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