# What is the molarity of the diluted solution when 1.00 L of a 0.250-M solution of Fe(NO3)3 is...

## Question:

What is the molarity of the diluted solution when 1.00 L of a 0.250-M solution of {eq}Fe(NO_3)_3 {/eq} is diluted to a final volume of 2.00 L?

## Molarity:

The molarity is the indication of the total moles of a chemical solution in its unit volume. Commonly, the proportion within the number of moles, and the volume of a chemical solution is utilized to determine the molarity. In chemistry, the capital M is used to denote the molarity of a chemical solution.

Given data

• The initial volume of the solution is {eq}{{V}_{1}}=1\ \text{L} {/eq}
• The initial molarity of the solution is {eq}{{M}_{1}}=0.25\ \text{M} {/eq}
• The final volume of the solution is {eq}{{V}_{2}}=2\ \text{L} {/eq}

By using the following relation, the molarity of the diluted solution is calculated as,

{eq}\begin{align*} {{M}_{1}}{{V}_{1}} &={{M}_{2}}{{V}_{2}} \\ \Rightarrow \left( 0.25\ \text{M} \right)\left( 1\ \text{L} \right) &={{M}_{2}}\left( 2\ \text{L} \right) \\ \Rightarrow {{M}_{2}} &=\dfrac{\left( 0.25\ \text{M} \right)\left( 1\ \text{L} \right)}{\left( 2\ \text{L} \right)} \\ \Rightarrow {{M}_{2}} &=0.125\ \text{M} \\ \end{align*} {/eq}

Thus, the molarity of the diluted solution is {eq}{{M}_{2}}=0.125\ \text{M} {/eq}.