Question

How many grams of potassium nitrate (KNO3) are required to prepare 0.250L of a 0.700 M
solution (KNO3 = 101.102 g/mol)
O 12.5 g
O 31.2g
25.38
17.78

Answer 1

Answer:

17.78g

Explanation:

m/M = C×V

m/101.102= 0.7× 0.25

m= 101.102×0.7×0.25= 17.78g