How many grams of ammonia are present in 5.0 L of a 0.05 M solution?
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Answered by
3
Given:
Volume, V=5L
Molarity, M=0.05mol/L
as Molar mass of NH₃(ammonia) is 17g
We know that,
M=[m]÷[V(L)×M.M.]
⇒m=M×M.M.×V
∴ m=0.05mol/L×17g×5L
⇒m=4.25g
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Answer:
previous answer is absolutely correct answer.
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