How milliliters of a 9.0 M $H_2SO_$ solution are needed to make 0.35 L of a 3.5 M solution?

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Answer 1 · 2016-09-29T20:37:10

You will need $"140 mL"$ of $"9.0 M H"_2"SO"_4"$ to make $"350 mL"$ of a $"3.5 M"$ solution.

The unit for molarity (M) is $"moles of solute"/"liters of solution"="mol/L"$ .

When diluting a solution, the amount of solute remains constant, but the volume of the solution increases.

The formula for the dilution of a solution is given below:

$"M"_1"V"_1"=M"_2"V"_2"$ ,

where $"M"$ is molarity and $"V"$ is volume of the solution in liters (L).

Known
$"M"_1="9.0 M"="9.0 mol/L"$
$"M"_2="3.5 M"="3.5 mol/L"$
$"V"_2="0.35 L"$

Unknown
$"V"_1"$

Solution
Rearrange the dilution formula to isolate $"V"_1"$ . Substitute the known values into the equation and solve.

$"V"_1=("M"_2"V"_2)/("M"_1)$

$"V"_1=(3.5cancel"M"xx"0.35 L")/(9.0cancel"M")="0.14 L"$ rounded to two significant figures

$"V"_1=0.14 cancel"L"xx(1000 "mL")/(1 cancel"L")="140 mL"$ rounded to two significant figures

How milliliters of a 9.0 M H_2SO_H_2SO_ solution are needed to make 0.35 L of a 3.5 M solution?