Question

38%(w/w) solution of HCl has density equal to 1.2g$ml^{-1}$.The molarity and molality of solution are:-

Answer 1

Answer:

You know that your solution is 38% w/w hydrochloric 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

1.00

L

⋅

1000

mL

1

L

⋅

1.19 g

1

mL

=

1190 g

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

1190

g solution

⋅

38 g HCl

100

g 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.

452.2

g

⋅

1 mole

36.46

g

=

12.40 moles

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

C

=

n

V

=

12.40 moles

1.00 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 solut

Answer 2

Answer:

Its a pretty good question?

Explanation: