Question

Commercially available Hcl 38% by mass .calculate molality of the solution.

Answer 1

Here's how you can approach such problems.

Explanation:

You know that your solution is 38% w/whydrochloric acid. This means that every 100 g of solution will contain 38 g of acid.

To make the calculations easier, assume that you're dealing with a 1.00-L sample of stock solution.

Use the solution's density to determine what the mass of this sample would be

1.00L⋅1000mL1L⋅1.19 g1mL=1190 g

Now use the known percent concentration by mass to determine how many grams of hydrochloric acid you'd get

1190g solution⋅38 g HCl100g solution=452.2 g HCl

To determine the solution's molarity, use hydrochloric acid's molar mass - this will get you the number of moles of acid present in the sample.

452.2g⋅1 mole36.46g=12.40 moles

Since the sample has a volume of 1.00-L, the molarity will be

C=nV=12.40 moles1.00 L=12.4 M

To get the solution's molality, you need to know the mass of water. Since you know the mass of the solution and that of acid, you can write

msol=mwater+macid

mwater=msol−macid=1190−452.2=737.8 g

The molality will thus be - do not forget to convert the mass of water to kilograms!

b=nmwater=12.40 moles737.8⋅10−3kg=16.8 molal

Since hydrochloric acid, HCl, can only release one mole of protons per mole of acid in aqueous solution, the solution's normality will be equal to its molarity.

normality=Cfeq, where