Answer:
MOLARITY = 0.9554 mol/dm3
Explanation:
525 grams of SnF2 in 3.50 L of wate
To obtain the molarity, calculate the concentration in g/dm3 and then concentration in mol/dm3
Concentration in g/dm3 = 525 g/ 3.50 L
= 525g/ 3.50 dm3
= 150g/dm3 of SnF4
Concnetration in mol/dm3
concentration in mol/dm3 = concentration in g/dm3 / Relative molecular mass of SnF2
RMM of SnF2 = ( 119 + 19*2) = 157 g/mol
= 150g/dm3 / 157g/mol
= 0.9554 mol/dm3 of solution.
The molarity of the solution is therefore 0.9554mol/dm3
