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JAMB Chemistry 2019 Paper

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Question : 31

Total: 80

At 27°C, 58.5g of sodium chloride is present in 250cm

^{3}

of a solution. The solubility of sodium chloride at this temperature is?

(molar mass of sodium chloride = 111.0gmol

^{-1}

)

2.0 moldm $^{-3}$
0.25 moldm $^{-3}$
1.0 moldm $^{-3}$
0.5 moldm $^{-3}$

Solution:

Concept:

Solubility in

\mathrm{mol\,dm}^{-3}

is the amount of solute in moles dissolved in

1\,\mathrm{dm}^3

of solution.

Explanation:

Convert the volume from

\mathrm{cm}^3

to

\mathrm{dm}^3

:

250\,\mathrm{cm}^3 = 0.25\,\mathrm{dm}^3

.

Calculate the mass concentration:

\frac{58.5\,\mathrm{g}}{0.25\,\mathrm{dm}^3} = 234\,\mathrm{g\,dm}^{-3}

.

Convert mass concentration to molar concentration using the given molar mass of

111.0\,\mathrm{g\,mol}^{-1}

:

\frac{234\,\mathrm{g\,dm}^{-3}}{111.0\,\mathrm{g\,mol}^{-1}} = 2.108\,\mathrm{mol\,dm}^{-3}

.

This is approximately equal to

2.0\,\mathrm{mol\,dm}^{-3}

.

Answer:

A.

2.0\,\mathrm{mol\,dm}^{-3}

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