Question

The concentration of a water solution of `NaCl` is 2.48 m, and it contains 806 g of water. How much `NaCl` is in the solution?

Answer 1

`"117 g"`

Explanation:

This is a great example of how a molality practice problem looks like. Before doing any calculations, make sure that you have a clear understanding of what molality means.

As you know, molality is defined as moles of solute, which in your case is sodium chloride, divided by the mass of solvent, always expressed in kilograms!

`color(blue)("molality" = "moles of solute"/"kilograms of solvent")`

So, a `"1-molal"` solution will contain `"1 mole"` of solute for every `"1 kg"` of solvent.

Now, your solution is said to have a molality of `"2.48 molal"`. This means that you would get `2.48` moles of sodium chloride for every `"1 kg"` of solvent.

However, you are told that your solution contains less than `"1 kg"` of water, your solvent. More specifically, it contains

`806 color(red)(cancel(color(black)("g"))) * " 1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.806 kg"`

of water. This means that you can expect it to contain fewer than `2.48` moles. Indeed, you can plug in your values to find

`0.806 color(red)(cancel(color(black)("kg water"))) * overbrace("2.48 moles NaCl"/(1color(red)(cancel(color(black)("kg water")))))^(color(purple)("the given molality")) = "1.9989 moles NaCl"`

Now, if you need to express this in grams of sodium chloride, simply use the compound's molar mass to go from moles to grams

`1.9989 color(red)(cancel(color(black)("moles NaCl"))) * "58.443 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = "116.82 g"`

Rounded to three sig figs, the answer will be

`m_(NaCl) = color(green)("117 g")`