Question #5998a
1 Answer
Answer:
Explanation:
We're asked to find (a) the molarity of the
(a)
To find the molarity of the solution, we can first recognize that in an

#80.0# #"g H"_2"SO"_4# 
#20.0# #"g H"_2"O"#
Also in a
#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
#80.0cancel("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
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 = (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"" "))#