# 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?

##### 1 Answer

#### 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 **for every**

Now, your solution is said to have a molality of **for every**

However, you are told that your solution contains **less** than

#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

#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")#