# Question 17171

Jul 27, 2015

Molarity: 9.9 M

#### Explanation:

In order to determine the molarity of a 31% w/w hydrochloric acid (muriatic acid) solution, you need to know its density.

At ${20}^{\circ} \text{C}$, the density of your solution is about $\text{1.16 kg/L}$. (http://www.handymath.com/cgi-bin/hcltble3.cgi?submit=Entry)

To make the calculations easier, use a 1.0-L sample of solution.

A solution's percent concentration by mass is defined as the mass of the solute, in your case hydrochloric acid, divided by the total mass of the solution, and multiplied by 100.

$\text{% w/w" = "mass of solute"/"mass of solution} \cdot 100$

So, you know that your solution is 31% w/w, which means that you get 31 g of solute, in your case hydrochloric acid, for every 100 g of solution.

Since you know what the density of the solution is, use its volume to determine its mass

1.0color(red)(cancel(color(black)("L"))) * "1.16 kg"/(1.0color(red)(cancel(color(black)("L")))) = "1.16 kg"

Expressed in grams, this is equal to

1.16color(red)(cancel(color(black)("kg"))) * "1000 g"/(1color(red)(cancel(color(black)("kg")))) = "1160 g"

The mass of hydrochloric acid you get from this solution is

1160color(red)(cancel(color(black)("g solution"))) * "31 g HCl"/(100color(red)(cancel(color(black)("g solution")))) = "359.6 g HCl"

Since molarity is defined as moles of solute per liters of solution, use hydrochloric acid's molar mass to determine how many moles you have

359.6color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.461color(red)(cancel(color(black)("g")))) = "9.86 moles HCl"

Therefore, the solution's molarity will be

C = n/V = "9.86 moles"/"1.0 L" = color(green)("9.9 M")#