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

Answer:

 $M_2  = 1.094 M$

Explanation:

Given data:

V1 = 25.0 ml

M2 = 0.00383 M

V2 = 1000 ml

we knwo that $M_1 V_1 = M_2 V_2$

    ∴ $M_1 = \frac{M_2 V_ 2}{V_1}$

              $= \frac{0.00383 M \times 1000}{25.0}$

               = 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

$M_2 = \frac{M_1 V_1}{V_2}$

     $= \frac{0.1532 M \times 125.0}{17.5}$

 $M_2  = 1.094 M$