If 50 cm3 3 of a saturated solution of KNO3 3 at 40 °C contained 5:05 g of the salt, its solubility at the same temperature would be [KNO3 3 = 101]

Answer Details

V= 50cm3 ${}^3$
Mass= 5.05g
Relative molecular mass of KNO3 ${}_3$ = (39+14+(3*16)) = 101
Convert 50cm3 ${}^3$ to dm3 ${}^3$ which is
1000cm³ = 1dm3 ${}^3$
50cm³ = 50*1/1000
= 0.05dm3 ${}^3$
Moles = mass/ molar mass
= 5.05/101 =0.05mole
Solubility= mole/volume
Solubility=0.05mol/0.05dm3 ${}^3$
Solubility=1.0mol/dm_3