Percent concentration by mass is defined as the mass of the solute, which in your case is "HCl", divided by the total mass of the solution and multiplied by 100.
The easiest way to approach such solution problems is by picking a convenient sample of the solution to base the calculations on. In this case, let's pick a "1-L" sample of your "38% concentration by mass solution.
The first thing you need to do is determine is how much the "1-L" sample weighs. Since density was given to you "g/cm"^3, it'll be best to transform it in "g/dm"^3, since "1 L = 1 dm"^3
1.19 "g"/"cm"^3 * ("1000 cm"^3)/("1 dm"^3) = "1190 g/dm"^3
Now focus on finding out how much "HCl" you have in this much solution.
#"38%" = m_("solute")/m_("solution") * 100 => m_("solute") = (m_("solution") * 38)/100#
m_("solute") = (38 * 1190)/100 = "452.2 g HCl"
For molarity, you need moles of solute per liter of solution. Use "HCl"'s molar mass to determine how many moles you have
"452.2 g" * ("1 mole")/("36.5 g") = "12.40 moles HCl"
Therefore,
C = n/V = "12.40 moles"/"1 L" = "12.4 M"
Molality will be moles of solute per kilogram of solution, so
"b" = n_("solute")/m_("solution") = ("12.40 moles")/(1190 * 10^(-3)"kg") = "10.4 molal"
For mole fraction you first need to determine the total number of moles you have in the sample. Find the number of moles of water by
m_("water") = "1190 g" - "452.2 g" = "737.8 g"
"737.8 g" * ("1 mole")/("18.0 g") = "41.0 moles water"
The total number of moles will be
n_("total") = n_("water") + n_("solute") = 41.0 + 12.40 = "53.4 moles"
Therefore, the mole fraction for "HCl" is
"mole fraction" = n_("solute")/n_("total") = 0.232
