Question

How many grams of 70% concentrated nitric acid solution should be used to prepare 250 mL of 2.0M?
A. $45.0g\;conc.\;HN{O_3}$
B. $90.0g\;conc.\;HN{O_3}$
C. $70.0g\;conc.\;HN{O_3}$
D. $54.0g\;conc.\;HN{O_3}$

Answer 1

If we know what is molarity, we can easily find the solution for this question. Molarity is the number of moles of solute dissolved per litre of the solution.

Complete step by step solution:
In question it is given that nitric acid has a molarity of 2.0.
So from the definition of molarity, we can get the number of moles of solute present in it.
$Molarity\;\left( M \right) = \dfrac{{Number\;of\;moles\;of\;solute}}{{Volume\;of\;solution\left( L \right)}}$
We have given that we used a solution having 250ml.
i.e.$Volume\; = \;250ml\; = 0.25L$
Therefore, we now know the volume and molarity of the solution. Now to find the number of moles of solute.
$Number\;of\;moles = Molarity \times Volume$
Substituting the values in the equation, we get
$Number\;of\;moles = 2.0 \times 0.25 = 0.5\;moles$
Therefore, the number of moles of solute in solution is 0.5
We know that 1 mole of $HN{O_3}$ contains $1g$ of Hydrogen, $14g$ of Nitrogen, $3 \times 16g$ of Oxygen. So the total mass of $HN{O_3}$ becomes $63g$.
That is 1 mole of $HN{O_3}$ contains $63g$. But we have only 0.5 moles
i.e, $mass\;of\;HN{O_3} = 0.5 \times 63 = 31.5g$
From question it is given that it is 70% concentrated, which means when there is 100g of solution it will have 70g of $HN{O_3}$
i.e, 70g of $HN{O_3}$ are present in the 100g of solution.
So, 1g of $HN{O_3}$ will be present in $\dfrac{{100}}{{70}}g$ of solution
Now, $31.5g$ of $HN{O_3}$ is present.
Then the mass of solution = $31.5 \times \dfrac{{100}}{{70}}$
$= 45g$
The mass of solution is $45g$.

**i.e., Option A is correct.

Note:**
We can use molarity for dilution of a solution
${M_1}{V_1} = {M_2}{V_2}$
Where ${M_1}$ is the initial molarity, ${V_1}$ is initial volume. ${M_2}$ is final molarity and ${V_2}$ is final volume.