Chemistry, asked by deathhero76, 8 months ago

The volume of 10.50M solution required to prepare 1.0 L of 0.25 M solution of Nitric acid is

Answers

Answered by Rajkd180203

0

Answer: 23.8 ml

M1V1=M2V2

10.50×V1=0.25×1.0

V1=0.25×1.010.50

=0.0238L=23.8 mL

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