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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##