Question

to 225ml of 0.80mol/L solution of kl, a student adds enough water to make 1.0L of a more dilute Kl solution what is the concentration of the new solution

Answer 1

Answer:

The concentration of the KI solution after diluting it with water is $0.18\ mol\ L^{-1}$·

Explanation:

Given that,

The initial volume of KI solution, $V_1$ $=$ $225$ ml $=\ 225$ × $10^{-3}$ L

The initial concentration of KI solution, $M_1$ $=\ 0.80\ \frac{mol}{L}$

The new volume after adding water, $V_2\ =\ 1.0$ L

To calculate the concentration of KI solution after diluting it with water we have the relationship that,

    $M_1\ V_1\ =\ M_2\ V_2$   $-----(1)$

where,

$M_1\ =$ The concentration of a solution in molarity

$V_1\ =$  The volume of the concentrated solution

$M_2\ =$ The concentration of the diluted solution

$V_2\ =$ The volume of the diluted solution

On rearranging the equation with the term $M_2$ on the left-hand side, we get

    $M_2\ =\ \frac{M_1\ V_1}{V_2}$  $-----(2)$

On substituting the values in equation No· $2$, we get

⇒ $M_2\ =\ \frac{0.80\ *\ 225\ *\ 10^{-3} }{1}$

⇒ $M_2\ =\ 0.18\ mol\ L^-^1$

Therefore,

The concentration of the KI solution after diluting with water is $0.18\ mol\ L^{-1}$·