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]
- 0.01 mol dm\(^{-3}\)
- 0.101 mol dm\(^{-3}\)
-
1.01 mol dm \(^{-3}\)
- 10.0 mol dm\(^{-3}\)
The correct answer is: C
Explanation
solubility( in mol/dm^3) = \(\frac{ mass of salt}{ Molar Mass of salt} = \frac{ 1000cm^3}{ volume of solution in cm^3}\)
\(\frac{14}{138 } = \frac{ 1000cm^3}{ 100}\)
\(\frac{14000}{13800} = 1.01 mol/dm^3\)