Question

Starting with a 6.847 M stock solution of HNO3, five standard solutions are prepared via serial dilution. At each stage, 25.00 mL of solution are diluted to 100.00 mL. Determine the concentration of and the number of moles of HNO3 in the final (most dilute, Md5) solution.

Answer 1

Answer: The concentration of
`HNO_3` in the final solution is 0.006688 M and number of moles are 0.00006688

Step-by-step explanation:

According to the neutralization law,

`M_1V_1=M_2V_2`

where,

`M_1` = molarity of stock solution = 6.847 M

`V_1` = volume of stock solution = 25.00 ml

`M_2` = molarity of ist dilute solution = ?

`V_2` = volume of first dilute solution = 100.0 ml

`6.847* 25.00=M_2* 100.0`

`M_2=1.712M`

2) on second dilution;

`1.712* 25.00=M_2* 100.0`

`M_2=0.4280M`

3) on third dilution

`0.4280* 25.00=M_2* 100.0`

`M_2=0.1070M`

4) on fourth dilution

`0.1070* 25.00=M_2* 100.0`

`M_2=0.02675M`

5) on fifth dilution

`0.02675* 25.00=M_2* 100.0`

`M_2=0.006688M`

Thus the concentration of
`HNO_3` in the final solution is 0.006688 M

moles of
`HNO_3` =
`Molarity* {\text {Volume in L}}=0.006688* 0.01L=0.00006688moles`