#"60.0 mL"# of a #"6.07 M"# stock solution of sulfuric acid is diluted to #"1.42 L"#. What is the molarity of the diluted solution?

1 Answer
Jun 1, 2017

Answer:

The molarity of the diluted sulfuric acid solution will be #"0.256 mol/L"#, or #"0.256 M"#.

Explanation:

When diluting a solution, you can determine the molarity of the diluted solution using the formula:

#M_1V_1=M_2V_2#,

where #M# is molarity in #"mol/L"# and #V# is the volume of the solution in liters.

The initial volume is given in #"mL"#, but it needs to be in liters. Convert #"60.0 mL"# into liters. #"1 L = 1000 mL"#

#60.0color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.0600 L"#

Organize your data.

Known

#M_1="6.07 M"="6.07 mol/L"#

#V_1="0.0600 L"#

#V_2="1.42 L"#

Unknown

#M_2=?#

Solution

Rearrange the formula above to isolate #M_2#. Insert the known data into the equation and solve.

#M_2=(M_1V_1)/(V_2)#

#M_2=((6.07"mol")/("1L")xx0.0600color(red)cancel(color(black)("L")))/(1.42color(red)cancel(color(black)("L")))="0.256 mol/L"# or #"0.256 M"# (rounded to three sig figs)