If 50 cm\(^{3}\) of a saturated solution of KNO\(_{3}\) at 40 °C contained 5:05 g of the salt, its solubility at the same temperature would be

[KNO\(_{3}\) = 101]

  • A 1.5 mol dm\(^{-3}\)
  • B 1.5 mol dm\(^{-3}\)
  • C 2.0 mol dm\(^{-3}\)
  • D 5.0 mol dm\(^{-3}\)
waec 2022chemistry

The correct answer is A. 1.5 mol dm\(^{-3}\)

V= 50cm\(^{3}\)

Mass= 5.05g

Relative molecular mass of KNO\(_{3}\) = (39+14+(3*16)) = 101

Convert 50cm\(^{3}\) to dm\(^{3}\) which is

1000cm³ = 1dm\(^{3}\)

50cm³ = 50*1/1000

= 0.05dm\(^{3}\)

Moles = mass/ molar mass

= 5.05/101 =0.05mole

Solubility= mole/volume

Solubility=0.05mol/0.05dm\(^{3}\)

Solubility=1.0mol/dm_\(^{3}\)

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