1.
Let's look at the problem: the "HCl" in solution is 38% by mass. If we assume a 100 "g" sample, then then the mass of "HCl" in solution is 38"g". The number of moles of the solute is thus
38cancel("g HCl")((1 "mol HCl")/(36.46cancel( "g HCl"))) = 1.04 "mol HCl"
Now, let's use the density of the solution (1.19 "g"/"mL"), and the fact that we assumed a 100 "g" sample to calculate the volume, in "L":
100 cancel("g")((1cancel( "mL"))/(1.19 cancel("g")))((1 "L")/(10^3 cancel("mL"))) = 0.0840 "L soln"
The molarity of the solution is thus
M = "mol solute"/"L soln" = (1.04 "mol HCl")/(0.0840 "L soln") = color(red)(12"mol"/"L"
2.
In 1000"mL" (or 1 "L") of a 1 M solution, the number of moles of "HCl" is
1 "mol"/cancel("L")(1 cancel("L")) = 1 "mol HCl"
We'll use this number and the fact that 1.04 "mol HCl" occupies a volume of 0.0840 "L" to find the volume of solution needed:
1 cancel("mol HCl")((0.0840 "L soln")/(1.04 cancel("mol HCl"))) = color(blue)(0.081 "L"), or color(blue)(81 "mL"
