Moles of chlorine gas needed to make up an equivalent amount of saturated salt solution

Question

I’m trying to solve a question about a saturated salt solution but am unsure of how to answer one of the parts or what it’s asking for.

The Problem

The maximum solubility of table salt (NaCl) in water at 25 degrees Celsius is approximately 359 g/L. Assume the volume of the solution does not change with the addition of salt.

a) What is the molar concentration of the mixture in mol/L?

b) What is the molal concentration of the mixture in mol/kg? Assume that the density of water is 980 kg/m³.

c) How many moles of Cl₂ gas would need to be present in order to make up an equivalent amount of saturated salt solution?

d) What is the density of the mixture?

My Attempt

a) molar concentration

359 g NaCl * 1 mol NaCl / 58.44 g NaCl = 6.14 mol => 6.14 mol/L

b) molal concentration

(1 L H₂O * 1 m³ H₂O * 980 kg) / (1000 L H₂O * 1 m³) = 6.14 mol / 0.980 kg = 6.26 mol/kg

c) moles of Cl₂ gas

I thought it was asking how many moles of chlorine gas is equivalent to the amount of salt in the 1 L solution (359 g NaCl). There are 70.9 g in 1 mole of Cl₂, so that gives me an answer of 5.06 mol Cl₂ gas (359 g NaCl / 70.9 g Cl₂).

But the answer key states the correct solution as 3.07 mol Cl₂ gas. How should I go about this part?

d) density of the mixture

density of water = 980 kg/m³

density of salt = 359 g/L => 359 kg/m³

density of mixture = (980 + 359) kg/m³ = 1339 kg/m³

Answer 1

a) molar concentration

There are indeed 6.14 moles of $\ce{NaCl}$ in 359 grams of $\ce{NaCl}$.

I'd prefer to write the molar mass as "$\pu{58.44 g mol^-1 }\ \ce{NaCl}$" or "$\pu{58.44 g/mol}\ \ce{NaCl}$" to keep it all together.

b) molal concentration

You got the right value, but you did it the wrong way. For the solvent:

$$\dfrac{\pu{980 kg/m^3}}{\pu{1000 L/m^3}} = \pu{0.980 kg/L}$$

c) moles of $\ce{Cl2}$ gas

The book is correct. There are 6.14 moles of $\ce{NaCl}$. Each molecule of chlorine gas contains two atoms of chlorine, so you need half as many moles to get the same number of atoms of Cl. So 6.14/2 = 3.07

This part of the question has poor phrasing. You can't dissolve 3.07 moles of $\ce{Cl2}$ in water and get a solution of $\ce{NaCl}$.

d) density of the mixture

Your answer is correct.