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

##### 1 Answer

#### 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,

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 **moles** of sodium chloride, your solute, in

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 **one mole** of sodium chloride has a mass of

In your case, **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.