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A 200 ml sample of 0.1015 Miric acid is mbred with 2300 ml of water. What is the molar concentration of nitric acid in the final solution 0 406 M 127M 325 x 10 M 5.08 x 10 M 8.12 x 10 M water Backspace lu 'o Pin

1 Answer

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Answer: Molar concentration of nitric acid in the final solution is
8.12* 10^(-3)M

Step-by-step explanation:

According to the dilution law,


M_1V_1=M_2V_2

where,


M_1 = molarity of stock
HNO_3 solution = 0.1015 M


V_1 = volume of stock
HNO_3solution = 200 ml


M_2 = molarity of dilute
HNO_3 solution = ?


V_2 = volume of dilute
HNO_3 solution = (2300 +200 )ml = 2500 ml

Putting in the values we get:


0.1015M* 200=M_2* 2500


M_2=8.12* 10^(-3)M

Thus the molar concentration of nitric acid in the final solution is
8.12* 10^(-3)M

User Hemant Kabra
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