Chemistry: A17.55 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.

A17.55 ml solution of potassium nitrate (kno3)

daevinepetersopa86on

11.01.2022

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

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