Question

A 17.55 mL solution of potassium nitrate (KNO3) was diluted to 125.0 mL, and 25.00 mL of this solution was then diluted to 1.000 × 103 mL. The concentration of the final solution is 0.00383 M. Calculate the concentration of the original solution.

Answer 1

`M_2  = 1.094 M`

Step-by-step explanation:

Given data:

V1 = 25.0 ml

M2 = 0.00383 M

V2 = 1000 ml

we knwo that
`M_1 V_1 = M_2 V_2`

∴
`M_1 = (M_2 V_ 2)/(V_1)`

`= (0.00383 M * 1000)/(25.0)`

= 0.1532 M

0.1532 M is concentration in 25 ml but taken from 125 ml solution.

∴ The concentration in 125.0 ml solution is = 0.1532 M

M1 = 0.1532 M

V1 = 125.0ml and V2 = 17.5.0ml

`M_2 = (M_1 V_1)/(V_2)`

`= (0.1532 M * 125.0)/(17.5)`

`M_2  = 1.094 M`