# How many grams of NaCl are required to prepare 985 mL of 0.77 M NaCl solution?

Jul 13, 2016

$\text{44 g NaCl}$

#### Explanation:

The problem provides you with the molarity and volume of the target solution, so your first step here will be to use this information to figure out how many moles of sodium chloride, $\text{NaCl}$, it must contain.

Once you know that, use the compound's molar mass to convert the number of moles to grams.

So, molarity is defined as the number of moles of solute per liter of solution. In your case, a $\text{0.77 M}$ solution will contain $0.77$ moles of sodium chloride, your solute, in $\text{1.0 L}$ of solution.

Your target solution has a volume of

985 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = "0.985 L"

which means that it will contain

0.985 color(red)(cancel(color(black)("L solution"))) * overbrace("0.77 moles NaCl"/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)("= 0.77 M")) = "0.75845 moles NaCl"

Sodium chloride has a molar mass of ${\text{58.44 g mol}}^{- 1}$, which basically means that one mole of sodium chloride has a mass of $\text{58.44 g}$.

In your case, $0.75845$ moles of sodium chloride will have a mass of

0.75845 color(red)(cancel(color(black)("moles NaCl"))) * "58.44 g"/(1color(red)(cancel(color(black)("mole NaCl")))) = color(green)(|bar(ul(color(white)(a/a)color(black)("44 g")color(white)(a/a)|)))

The answer is rounded to two sig figs, the number of sig figs you have for the molarity of the solution.