# The molarity of a solution composed of 25.4g {eq}KHCO_3 {/eq} diluted with water to a total volume of 200 mL.

## Question:

The molarity of a solution composed of 25.4g {eq}KHCO_3 {/eq} diluted with water to a total volume of 200 mL.

## Molarity of a Solution

The number of moles of solute dissolved in a solution which has a volume of one liter, then this measurement for defining concentration term is called molarity of the solution.

We are given:

• The given mass of potassium bicarbonate in the solution = 25.4 g
• The volume of the given solution = 200 mL = 0.2 L

We know:

• The molar mass of potassium bicarbonate = 100 g/mol

Let us find the number of moles of potassium bicarbonate in the solution.

The number of moles of potassium bicarbonate = {eq}\dfrac{25.4\ g}{100\ g/mol}= 0.254\ mol {/eq}

We know that

{eq}\begin{align} \textrm{Molarity of the solution}&=\dfrac{\textrm{Number of moles of potassium bicarbonate}}{\textrm{Volume of solution in liter}}\\ &= \dfrac{0.254\ mol}{0.2\ L}\\ &= 1.27\ M \end{align} {/eq}

Hence, the molarity of the solution is 1.27 M.