Question

What is the molarity of a salt solution made by dissolving 240mg of NaCl in 4ml of water? Assume final volume is the same as the volume of water.

Answer 1

`"1 mol L"^(-1)`

Explanation:

In order to find the molarity of the solution, you must determine the number of moles of solute present in exactly

`"1 L" = 10^3` `"mL"`

of solution. Notice that you already know the number of milligrams of sodium chloride, the solute, present in `"4 mL"` of solution, so start by calculating the mass of solute present in `10^3` `"mL"` of solution.

`10^3 color(red)(cancel(color(black)("mL solution"))) * "240 mg NaCl"/(4color(red)(cancel(color(black)("mL solution")))) = 60 * 10^3` `"mg"`

Since

`color(blue)("1 g" = 10^3color(white)(.)"mg"`

you can say that `10^3` `"mL"` of solution contain

`60 * color(blue)(10^3color(white)(.)"mg") = 60color(white)(.)color(blue)("g")`

of sodium chloride. Now, to convert this to moles, use the molar mass of the compound

`60 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.44color(red)(cancel(color(black)("g")))) = "1.027 moles NaCl"`

Since this represents the number of moles of sodium chloride present in `10^3` `"mL"` of solution, you can say that the molarity of the solution is equal to

`color(darkgreen)(ul(color(black)("molarity = 1 mol L"^(-1)))`

The answer must be rounded to one significant figure, the number of sig figs you have for the volume of the solution.

Answer 2

M=`"1 mol/L"`

Explanation:

The molarity of a solution is the concentration of moles of solute per liter of solution, `"mol/L"`. The mass of `"NaCl"` must be converted to grams and then moles. The volume of the solution must be converted to liters.

Convert `"mg NaCl"` to `"g NaCl"`.

`"1 g=1000 mg"`

`240color(red)cancel(color(black)("mg NaCl"))xx(1"g")/(1000color(red)cancel(color(black)("mg")))="0.24 g NaCl"`

Determine moles of `"NaCl"`.

Divide the mass of `"NaCl"` by its molar mass by multiplying by its inverse.

Molar mass `"NaCl=58.44 g/mol"`

`0.24color(red)cancel(color(black)("g NaCl"))xx(1"mol NaCl")/(58.44color(red)cancel(color(black)("g NaCl")))="0.0041 mol NaCl"`

Molarity of `"NaCl"`.

Convert `"4 mL"` to liters.

`"1 L=1000 mL"`

`4color(red)cancel(color(black)("mL"))xx(1"L")/(1000color(red)cancel(color(black)("mL")))="0.004 L"`

Divide mol NaCl by volume of solution.

`"M"=(0.0041"mol NaCl")/(0.004"L solution")="1 mol/L"` rounded to one sig fig due to `4` mL