To what volume must $50 m l$ of $3.50 M H_{2} S$ be diuted in order to make $2 M H_{2} S O_{4}$ ?

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Solution

$C_{1} V_{1} = C_{2} V_{2}$

volume = $50$ ml or $0.050$ litres
Concentration = $3.50 M$ or $3.50 m o l e s / l i t r e$
New volume = ?
New concentration = $2.00 M$ or $2.00 m o l e s / l i t r e$

$C_{1} V_{1} = C_{2} V_{2}$

$V_{2} = \frac{C_{1} V_{1}}{C_{2}}$

$V_{2} = \frac{350 M \times 0.050 L i t r e}{2.0 m} = 0.0875 L i t r e$
This makes sense since the diluted volume will have more amount of water and thus the new volume diluted will be more than $V_{1}$ in the original solution

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