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

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

$6.847\times 25.00=M_2\times 100.0$

$M_2=1.712M$

2) on second dilution;

$1.712\times 25.00=M_2\times 100.0$

$M_2=0.4280M$

3) on third dilution

$0.4280\times 25.00=M_2\times 100.0$

$M_2=0.1070M$

4) on fourth dilution

$0.1070\times 25.00=M_2\times 100.0$

$M_2=0.02675M$

5) on fifth dilution

$0.02675\times 25.00=M_2\times 100.0$

$M_2=0.006688M$

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

moles of $HNO_3$ = $Molarity\times {\text {Volume in L}}=0.006688\times 0.01L=0.00006688moles$