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QUESTION

# Density HCl 1.19 g/cm^3 Mass Percent of Solute 38 what is the molarity, molality, and mole fraction?

is defined as the mass of the , 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 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 , 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"

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, fraction for "HCl" is

"mole fraction" = n_("solute")/n_("total") = 0.232