Question

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

Answer 1

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_3`solution = 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`