Question

calculate the molaity of a solution made by dissolving 40g NaCl in 300g of water

Answer 1

Task:

calculate the molality of a solution made by dissolving 40g NaCl in 300g of water

Solution:

It's not clear what we have to find in the task. Is "molality" a molarity or molality?

So let's find both of them.

1) Molarity

$$C(M) = n(\text{mol}) / V(L)$$

C – molarity of the solution (mol/L)

n – number of moles of NaCl

V – volume of solution (L)

The amount of NaCl is

$$n(\text{mol}) = m(g) / MW(g/\text{mol})$$ $$MW(\text{NaCl}) = MW(\text{Na}) + MW(\text{Cl}) = 23 + 35.5 = 58.5\ \text{g/mol}$$ $$n(\text{NaCl}) = 40 / 58.5 = 0.684\ \text{mol}$$

The density of water $d = 1.00\ \text{g/cm}^3$

The volume of solution $\approx$ the volume of water

$$V(\text{water}) = m(g) / d(g/\text{cm}^3) = 300 / 1.00 = 300\ \text{mL} = 0.300\ \text{L}$$

The molarity of NaCl solution is

$$C(\text{NaCl}) = n(\text{NaCl}) / V(L)$$ $$C(\text{NaCl}) = 0.684 / 0.300 = 2.28\ \text{M}$$

2) Molality – number of moles of substance in $1\ \text{kg}$ of solvent

$$M(\text{mol} / \text{kg}) = n(\text{mol}) / m(\text{kg})$$

The number of moles of NaCl we have already found

$$n(\text{NaCl}) = 0.684\ \text{mol}$$ $$m(H_2O) = 300\ \text{g} = 0.300\ \text{kg}$$

The molality of the solution is

$$M(\text{mol} / \text{kg}) = n(\text{mol}) / m(\text{kg})$$ $$M(\text{NaCl}) = n(\text{NaCl}) / m(\text{H}_2\text{O})$$ $$M(\text{NaCl}) = 0.684 / 0.300 = 2.28\ \text{mol/kg}$$

In our case the molarity is equal to molality

Answer: $C(\text{NaCl}) = 2.28\ \text{M}$; $M(\text{NaCl}) = 2.28\ \text{mol/kg}$