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]

1.0 mol dm $^{-3}$

1.5 mol dm $^{-3}$

2.0 mol dm $^{-3}$

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_ $^{3}$

See all Chemistry questions & answers

See more WAEC questions & answers

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]

Similar Questions

Aptitude Tests

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]

Similar Questions

Aptitude Tests

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]