Answer to Question #298422 in General Chemistry for Mitsu

Question

calculate the molality and molarity of aqueous phosphoric acid solution 1.88 gram/ml containing 20% by mass of h3 po4? h=1 p=31,o=16

Answer 1

Explanation:

As you know, molality is defined as moles of solute, which in your case is phosphoric acid, H

3

PO

4

, divided by kilograms of solvent, which is of course water.

b

=

n

solute

m

solvent

This means that you need to use the solution's percent concentration by mass to determine how much solute and how much solvent you get in a random sample of solution.

Since you're dealing with percentages, you can make the calculations easier by picking a 100.0-g

 sample of this solution. According to the given percent concentration, this solution will contain 21.6 g

 of phosphoric acid.

Remember, percent concentration by mass is calculated using the mass of solute and the mass of the solution, which includes both the solute and the solvent.

This means that the sample will also contain

m

sample

=

m

solute

=

m

solvent

m

water

=

100.0 g

−

21.6 g

=

78.4 g

Do not forget that you need to convert the mass of the solvent from grams to kilograms by using the conversion factor

1 kg

=

10

3

g

78.4

g

⋅

1 kg

10

3

g

=

78.4

⋅

10

−

3

g

Now, to find the number of moles of phosphoric acid present in this sample, use the compound's molar mass.

21.6

g

⋅

1 mole H

3

PO

4

97.995

g

=

0.2204 moles H

3

PO

4

Therefore, the solution's molality will be

b

=

0.2204 moles

78.4

⋅

10

−

3

kg

=

2.81 molal