The 25°C evaporation of a 100 cm\(^{3}\) solution of K\(_{2}\)CO\(_{3}\) to dryness gave 14g of the salt. What is the solubility of K\(_{2}\)CO\(_{3}\) at 25°C? [K\(_{2}\)CO\(_{3}\) = 138]

A.

0.01 mol dm\(^{-3}\)

B.

0.101 mol dm\(^{-3}\)

C.

1.01 mol dm \(^{-3}\)

D.

10.0 mol dm\(^{-3}\)

Correct answer is C

solubility( in mol/dm^3) = \(\frac{ mass of salt}{ Molar Mass of salt} = \frac{ 1000cm^3}{ volume of solution in cm^3}\)

\(\frac{14}{138 } = \frac{ 1000cm^3}{ 100}\)

\(\frac{14000}{13800} = 1.01 mol/dm^3\)

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