Question #5998a

We're asked to find (a) the molarity of the #"H"_2"SO"_4# solution, and (b) the volume of that acid solution required to make #1# #"L"# of a #0.2# #M# solution.

(a)

To find the molarity of the solution, we can first recognize that in an #80%# by mass solution, and we assume a #100#-#"g"# sample, there would be

• #80.0# #"g H"_2"SO"_4#

• #20.0# #"g H"_2"O"#

Also in a #100#-#"g"# sample, using the given density, the volume (in liters) of the solution is

#V = 100cancel("g soln")((1cancel("mL soln"))/(1.787cancel("g soln")))((1color(white)(l)"L soln")/(10^3cancel("mL soln"))) = color(red)(ul(0.0560color(white)(l)"L soln"#

Remember that the molarity equation is

#ul("molarity" = "mol solute"/"L soln"#

We just found the liters of solution, so let's use the #80# #"g H"_2"SO"_4# and the molar mass of #"H"_2"SO"_4# to find the moles:

#80.0cancel("g H"_2"SO"_4)((1color(white)(l)"mol H"_2"SO"_4)/(98.078cancel("g H"_2"SO"_4))) = color(green)(ul(0.816color(white)(l)"mol H"_2"SO"_4#

The molarity of the solution is thus

#"molarity" = color(green)(0.816color(white)(l)"mol H"_2"SO"_4)/(color(red)(0.0560color(white)(l)"L soln")) = color(blue)(ulbar(|stackrel(" ")(" "14.6color(white)(l)M" ")|)#

(b)

Now, we can use the dilution equation to find the volume of this solution required to make #1# #"L"# of a #0.2# #M# solution.

The dilution equation is

#ul(M_1V_1 = M_2V_2#

where

• #M_1# and #M_2# are the molarities of the two solutions

• #V_1# and #V_2# are the respective volumes of the two solutions

We know:

• #M_1 = color(blue)(14.6color(white)(l)M#

• #V_1 = ?#

• #M_2 = 0.2color(white)(l)M#

• #V_2 = 1color(white)(l)"L"#

Let's rearrange the equation to solve for the unknown volume, #V_1#:

#V_1 = (M_2V_2)/(M_1)#

Plugging in known values:

#V_1 = ((0.2cancel(M))(1color(white)(l)"L"))/(color(blue)(14.6)cancel(color(blue)(M))) = color(blue)(0.0137color(white)(l)"L"# #= color(blue)(ulbar(|stackrel(" ")(" "13.7color(white)(l)"mL"" ")|)#