chemistry question

Question

Calculate the molality of each of the following aqueous solutions:

  (a) 2.50 M NaCl solutions (density of the solution=1.08 g/mL)

        Answer=2.68 m

  (b) 48.2 percent by mass KBr solution.

       Answer= 7.82 m

what's the equation for it? Because I have the answers

Answer 1

There really are no specific equations, but rather a series of operations. If you understand what molal means, and what molar means, you can find the answers.

Molality = moles solute / kg solvent

a). Assume we have 1000 ml of solution.

1000 ml x 1.08 g/ml = 1080 g

Now find mass of NaCl:

2.50 M NaCl = 2.50 mol/L and 2.50 mol/L x 1 L = 2.50 moles NaCl

molar mass NaCl = 58.4 g/mol, therefore mass NaCl present = 2.50 mol x 58.5 g/mol = 146 g NaCl

Now find mass of H2O:

grams H₂O present = 1080 g - 146 g = 934 g H₂O = 0.0.934 kg

We now have moles NaCl and kg H₂O

Molality = 2.50 mol / 0.934kg = 2.68 molal

b). 48.2% by mass KBr means 48.2 g KBr / 100 g solution

To find molality, we need moles of KBr and kg of H₂O. We can easily find moles KBr from the mass and the molar mass. But how do we find mass of H₂O?

If we have 100 g solution, 48.2 g is KBr so the rest is H₂O.

100 g - 48.2 g = 51.8 g H₂O and this is equal to 0.0518 kg H₂O

Moles KBr = 48.2 g KBr x 1 mol KBr/119 g = 0.405 moles KBr

Molality = 0.405 moles KBr / 0.0518 kg H₂O = 7.82 m