# Question cbe8e

Jan 22, 2016

${M}_{1} = 2.5 M$

#### Explanation:

Since the number of mole remains the same, we can then say:

${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$

Where, M_1=?#, ${V}_{1} = 10 m L$, ${M}_{2} = 0.1 M$ and ${V}_{2} = 250 m L$.

The concentration of the stock solution is then found by:

$\implies {M}_{1} = \frac{{M}_{2} {V}_{2}}{{V}_{1}} = \frac{0.1 M \times 250 \cancel{m L}}{10 \cancel{m L}} = 2.5 M$