Question

Calculate [NO3-] if 125 mL of 0.35 M NaNO3 is mixed with 450 mL of 1.1 M Mg(NO3)2. Please include some of your work as best as you can in the answer for full marks.

Answer 1

Answer: 1.76 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$     .....(1)

Molarity of $NaNO_3$ solution = 0.35 M

Volume of solution = 125 mL

Putting values in equation 1, we get:

a) $0.35M=\frac{\text{Moles of}NaNO_3\times 1000}{125ml}\\\\\text{Moles of }NaNO_3=\frac{0.35mol/L\times 125}{1000}=0.044mol$

1 mole of $NaNO_3$ contains = 1 mol of $NO_3^-$

Thus $0.044mol$ of $NaNO_3$ contain= $\frac{1}{1}\times 0.044=0.044$ mol of $NO_3^-$

b) $1.1M=\frac{\text{Moles of}Mg(NO_3)_2\times 1000}{450ml}\\\\\text{Moles of }Mg(NO_3)_2=\frac{1.1mol/L\times 450}{1000}=0.495mol$

1 mole of $Mg(NO_3)_2$ contains = 2 mol of $NO_3^-$

Thus $0.495mol$ of $Mg(NO_3)_2$ contain= $\frac{2}{1}\times 0.495=0.99$ mol of $NO_3^-$

Total $[NO_3^-]=\frac {\text {total moles}}{\text {total volume}}=\frac{0.044+0.99}{0.575L}=1.76M$

Thus $[NO_3^-$ after mixing is 1.76 M