Question

Determine the molarity of a solution made by dissolving 11.7 g of NaNO3 in water where the final volume of the solution is 250.0 mL.

Answer 1

`0.552~M`

Step-by-step explanation:

For the calculation of molarity "M" we have start with the molarity equation:

`M=(mol)/(L)`

So, we have to calculate the moles of
`NaNO_3` and the L of
`NaNO_3`.

For the calculations of moles we have to use the molar mass of
`NaNO_3`.

Na=23 g/mol

N=14 g/mol

O= 16 g/mol

`molar~mass~=~(23*1)+(14*1)+(16*3)=85~g/mol`

or

`1~mol~NaNO_3=85~g~NaNO_3`

Now, we can find the moles of
`NaNO_3`:

`11.7~g~NaNO_3*(1~mol~NaNO_3)/(85~g~NaNO_3) =0.138~mol~NaNO_3`

The next step would be the converstion from mL to L:

`250.0~mL~*(1~L)/(1000~mL) =~0.25~L\\`

Finally, we have to plug both values in the molarity equation:

`M=(0.138~mol)/(0.25~L)=~0.552~M`

Answer 2

Molarity is defined as the number of moles of solute in 1 L solution.
we have been given the mass of NaNO₃ dissolved in 250.0 mL
number of moles of NaNO₃ - 11.7 g / 85 g/mol = 0.138 mol
number of NaNO₃ moles in 250.0 mL - 0.138 mol
since molarity is the number of moles in 1 L
number of NaNO₃ moles in 1000 mL - 0.138 mol / 250.0 mL x 1000mL/L = 0.552 mol
molarity of NaNO₃ is 0.552 M