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Paper 2 · November 2010 · Chemical Equilibrium

The iodine liberated in an experiment required 15.0 cm315.0\ cm^3 of 0.03 mol dm−30.03\ mol\,dm^{-3} sodium thiosulphate. The number of moles of thiosulphate used is

A4.50×10−44.50 \times 10^{-4}
B4.50×10−34.50 \times 10^{-3}
C4.50×10−24.50 \times 10^{-2}
D2.25×10−42.25 \times 10^{-4}

Explanation

n=cV=0.03×15.0/1000=4.5×10−4 moln = 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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