The process of reducing the concentration of solute in a solution from higher to lower is known as dilution. Adding more solvent into the solution, dilute the solution without adding solute. The dilution calculator helps to calculate the concentration of solution at initial and final dilution. The formula to calculate the initial and the final concentration of the solution is given by,

Where,

$m_{1}$ = initial concentration of the solution i.e., before dilution,

$m_{2}$ = final concentration of the solution i.e., after dilution,

$V_{1}$ = initial volume of the solution i.e., before dilution,

$V_{2}$ = final volume of the solution i.e., after dilution.

$m_{1}$ $\times$ $V_{1}$ = $m_{2}$ $\times$ $V_{2}$

Where,

$m_{1}$ = initial concentration of the solution i.e., before dilution,

$m_{2}$ = final concentration of the solution i.e., after dilution,

$V_{1}$ = initial volume of the solution i.e., before dilution,

$V_{2}$ = final volume of the solution i.e., after dilution.

**Step 1 :**Make a note of initial and final concentration and also volume of the given solution.

**Step 2 :**Using the above formula, substitute the values into the formula and get the appropriate value accordingly.

$m_{1}$ $\times$ $V_{1}$ = $m_{2}$ $\times$ $V_{2}$

Solved problems are given below based on dilution.

### Solved Examples

**Question 1:**A chemistry teacher start an experiment with 60.0 mL of a 0.40 M HCl solution and diluting it in 1000 mL. Determine the concentration of HCl in the new solution.

**Solution:**

Given : $m_{1}$ = 0.40 M;

$V_{1}$ = 60.0 mL = 0.06 L;

$V_{2}$ = 1000.0 mL = 1L;

$m_{2}$ = ?

Final concentration of HCl solution is given by,

$m_{2}$ = $\frac{m_{1} \times V_{1}}{V_{2}}$

$m_{2}$ = $\frac{0.40 \ M \times 0.06 \ L}{1 \ L}$

$m_{2}$ = 0.024 M

**Question 2:**Find the final volume of the solution that requires 0.55 M of solution and 200.0 mL of 2.5 M KCl solution is diluted.

**Solution:**

Given : $m_{1}$ = 2.5 M;

$V_{1}$ = 200.0 mL;

$m_{2}$ = 0.55 M;

$V_{2}$ = ?

Final volume of KCl solution is given by,

$V_{2}$ = $\frac{m_{1} \times V_{1}}{m_{2}}$

$V_{2}$ = $\frac{2.5 \ M \times 200.0 \ mL}{0.55 \ M}$

$V_{2}$ = 909.09 mL = 0.909 L