# Question #c92ff

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

**Molality** is defined as the number of moles of solute present **for every** **of solvent**.

You know that your solution has a *molarity* of **every** **of solution** contains **moles** of sodium chloride, the solute.

To make the calculations easier, let's pick a **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 **moles** of sodium chloride **for every** **of water**, the solvent, you can say that

#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.