Question

Question #c92ff

Answer 1

Here's what I got.

Explanation:

Molality is defined as the number of moles of solute present for every `"1 kg"` of solvent.

You know that your solution has a molarity of `"3 M"`, which implies that every `"1 L"` of solution contains `3` moles of sodium chloride, the solute.

To make the calculations easier, let's pick a `"1-L"` sample of this solution. Use the molar mass of sodium chloride to calculate how many grams of solute it contains

`3 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = "175.32 g"`

Now, use the density of the solution to determine its mass.

`1 color(red)(cancel(color(black)("L solution"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("1.25 g"/(1color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the density of the solution")) = "1250 g"`

To find the mass of water present in the sample, subtract the mass of the solute from the mass of the solution

`m_"water" = "1250 g" - "175.32 g"`

`m_"water" = "1074.68 g"`

This is equivalent to

`1074.68 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "1.07468 kg"`

So, if your solution contains `3` moles of sodium chloride for every `"1.07468 kg"` of water, the solvent, you can say that `"1 kg"` of water will contain

`1 color(red)(cancel(color(black)("kg water"))) * "3 moles NaCl"/(1.07468 color(red)(cancel(color(black)("kg water")))) = "2.79 moles NaCl"`

Therefore, the molality of the solution will be equal to

`color(darkgreen)(ul(color(black)("molality = 2.8 mol kg"^(-1))))`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the molarity of the solution.