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

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

Answer 2

Answer:

$0.552~M$

Explanation:

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

$M=\frac{mol}{L}$

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  $NaNO_3$:

$11.7~g~NaNO_3*\frac{1~mol~NaNO_3}{85~g~NaNO_3} =0.138~mol~NaNO_3$

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

$250.0~mL~*\frac{1~L}{1000~mL} =~0.25~L$

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

$M=\frac{0.138~mol}{0.25~L}=~0.552~M$