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

asked Dec 4, 2021 29.0k views

1 vote

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

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2 votes

Answer:

17.78g

Step-by-step explanation:

m/M = C×V

m/101.102= 0.7× 0.25

m= 101.102×0.7×0.25= 17.78g

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