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

Question

Avery Adams · Accepted Answer

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

Jose Ayala · Answer

undefined You can use the dilution formula to solve this problem. The dilution formula is:

$ M_1V_1 = M_2V_2 $

Given:
$M_1 = 9.0 \, \text{M}$
$M_2 = 3.5 \, \text{M}$
$V_2 = 0.35 \, \text{L}$

Substituting the values into the dilution formula and solving for $V_1$:

$ (9.0 \, \text{M})(V_1) = (3.5 \, \text{M})(0.35 \, \text{L}) $

$ V_1 = \frac{(3.5 \, \text{M})(0.35 \, \text{L})}{9.0 \, \text{M}} $

$ V_1 ≈ 0.136 \, \text{L} $

Since $1 \, \text{L} = 1000 \, \text{mL}$, you can convert $0.136 \, \text{L}$ to milliliters:

$0.136 \, \text{L} \times 1000 \, \text{mL/L} = 136 \, \text{mL}$

Therefore, you need 136 milliliters of a 9.0 M H2SO4 solution to make 0.35 L of a 3.5 M solution.