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.
1.0 mol dm\(^{-3}\)
B.
1.5 mol dm\(^{-3}\)
C.
2.0 mol dm\(^{-3}\)
D.
5.0 mol dm\(^{-3}\)
Correct answer is A
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_\(^{3}\)
