Question

Commercial grade hcl solutions are typically 39.0% (by mass) hcl in water. determine the molarity of the hcl, if the solution has a density of 1.20 g/ml.

Answer 1

Answer:

12.82 mol/L the molarity of the HCl.

Explanation:

Suppose in 100 grams of 39.0% (by mass) HCl in water.

Volume of solution = V

Density of the solution = d = 1.20 g/mL

Mass = Density × Volume

$V=\frac{M}{d}=\frac{100 g}{1.20 g/mL}=83.33 mL = 0.08333 L$

Mass of HCl = 39.0% of 100 grams= $\frac{39}{100}\times 100g=39 g$

Moles of HCl = $\frac{39 g}{36.5 g/mol}=1.0685 mol$

$Molarity=\frac{\text{Moles of compound}}{\text{Volume of solution (L)}}$

The molarity of the HCl = M

$M=\frac{1.0685 mol}{0.0833 L}=12.82 mol/L$

12.82 mol/L the molarity of the HCl.

Answer 2

To make computations easier, let us assume that there is 1 liter solution of the sample.

So using the density formula to know what the mass of this sample would be:

1 L x 1000 mL / 1L x 1.20g/ 1 mL = 1120g

Now use the percent concentration by mass to know how many grams of HCl you can have:
1120 g sol x 39 g HCl / 100 g = 313.6 g HCL

To determine the solution's molarity, use the molar mass to know the number of moles present:

313.6g x 1 mole / 36.46g = 8.60 moles

since the sample has a volume of 1 L, the molarity is:

C = n / V = 8.60 moles / 1.0 L = 8.60 M

To get the solution's molarity, determine the mass of water.
Msol = Mwater + Macid
Mwater = Msol - Macid = 1120 - 313.6 = 806.4 g

The molarity will be:
b = n/ Mwater = 8.60 moles / 806.4 x 10^-3 = 10.66 molal

the solution's molarity will be equal to its normality since HCl acid can only release one mole of protons per mole of acid
Normality = 8.6 N