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](http://socratic.org/chemistry/solutions-and-their-behavior/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#
