Question #03a5a

1 Answer
Aug 5, 2017

Answer:

#color(red)"8.3 mL"# of a 6.0 M solution of HCl to make 100.0 mL of a 0.50 M solution of HCl.

Explanation:

This is a dilution problem involving molarity. The formula to use is:

#M_1V_1=M_2V_2#,

where #M# is molarity and #V# is volume in liters.

The unit for molarity is #"mol solute/L of solution"#, or simply #"mol/L"#. So a 2.5 M solution contains #"2.5 mol solute/L of solution"#. Therefore, all volumes must be in liters. The final volume can be converted into milliliters, as is the case in this question.

Organize the data:

Known

#M_1="6.0 M HCl"##=##"6.0 mol/L HCl"#

#M_2="0.50 M HCl"##=##"0.50 mol/L HCl"#

#V_2=100.0 color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.1000 L"#

Unknown

#V_1#

Solution

Rearrange the formula to isolate #V_1#. Insert the data into the equation and solve.

#V_1=(M_2V_2)/M_1#

#V_1=(0.50color(red)cancel(color(black)("M"))xx0.1000"L")/(6.0color(red)cancel(color(black)("M")))="0.0083 L"#

(rounded to two significant figures)

Convert liters to milliliters.

#0.0083color(red)cancel(color(black)("L"))xx(1000"mL")/(1color(red)cancel(color(black)("L")))="8.3 mL"#