1. Starting with a 0. 1525 m hcl stock solution, three standard solutions are prepared by sequentially diluting 5. 00 ml of each solution to 100. 0 ml. What is the concentration of each solution?.



Answer :

The concentration of each solution is 0.007625 M,  0.0003813 M, 0.00001906 M.

For first solution

Initial concentration = n1 = 0.1525 M

Volume taken = v1 = 5 ml

Final concentration = n2 = ?

Final volume = v2 = 100 ml

We know,

n1 v1 = n2 v2

n2 = n1v1/ v2 = ( 0.1525 M × 5ml) / 100 ml

n2 = 0.007625 M

Concentration of first solution = 0.007625 M

For 2nd solution

Initial concentration = n1 = 0.007625 M

Volume taken = v2 = 5 ml

Final concentration = n2 = ?

Final volume = v2 = 100 ml

We know,

n1 v1 = n2 v2

n2 = n1v1/ v2 = ( 0. 007625 M × 5ml) / 100 ml

n2 = 0.0003813 M

Concentration of second solution = 0.0003813 M

Fir third solution

Initial concentration = n1 = 0.0003813 M

Volume taken = v1 = 5 ml

Final concentration = n2 = ?

Final volume = v2 = 100 ml

We know,

n1 v1 = n2 v2

n2 = n1v1/ v2 = ( 0.0003813 M × 5ml) / 100 ml

n2 = 0.00001906 M

Concentration of first solution = 0.00001906 M

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