# Question eb6b6

Feb 6, 2015

Percent concentration by mass is defined as the mass of the solute, which in your case is $\text{HCl}$, 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 $\text{1-L}$ sample of your "38% concentration by mass solution.

The first thing you need to do is determine is how much the $\text{1-L}$ sample weighs. Since density was given to you ${\text{g/cm}}^{3}$, it'll be best to transform it in ${\text{g/dm}}^{3}$, since ${\text{1 L = 1 dm}}^{3}$

$1.19 {\text{g"/"cm"^3 * ("1000 cm"^3)/("1 dm"^3) = "1190 g/dm}}^{3}$

Now focus on finding out how much $\text{HCl}$ 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 $\text{HCl}$'s molar mass to determine how many moles you have

$\text{452.2 g" * ("1 mole")/("36.5 g") = "12.40 moles HCl}$

Therefore,

$C = \frac{n}{V} = \text{12.40 moles"/"1 L" = "12.4 M}$

Molality will be moles of solute per kilogram of solution, so

$\text{b" = n_("solute")/m_("solution") = ("12.40 moles")/(1190 * 10^(-3)"kg") = "10.4 molal}$

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"

$\text{737.8 g" * ("1 mole")/("18.0 g") = "41.0 moles water}$

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 $\text{HCl}$ is

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