If 50 cm3 of a saturated solution of KNO3 at 40 °C contained 5:05 g of the salt; its solubility at the same temperature would be [KNO3 = 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

Correct Answer:

Option A = 1.5 mol dm−3

Explanation

V= 50cm3
Mass= 5.05g
Relative molecular mass of KNO3 = (39+14+(3*16)) = 101
Convert 50cm3 to dm3 which is
1000cm³ = 1dm3
50cm³ = 50*1/1000
= 0.05dm3
Moles = mass/ molar mass
= 5.05/101 =0.05mole
Solubility= mole/volume
Solubility=0.05mol/0.05dm3
Solubility=1.0mol/dm_3