Question

How many grams of nitric acid are present in $250.0ml$ of $6.70M$ acid solution?

Answer 1

Hint: This question is based on the concept of molarity, if we are thorough with the term Molarity, we can easily attempt the question. Molarity is defined as the number of moles of solute dissolved per litre of the solution.

Complete step-by-step answer:
It is given in the question, that the molarity of the acid=$6.7M$
And volume of solution= $250ml\,or\,\dfrac{250}{1000}=\,0.25l$
Also, from the definition on molarity we know that
$Molarity\,\left( M \right)=\dfrac{Number\,of\,moles\,of\,solute}{volume\,of\,solution\,\left( in\,L \right)}$
From the above formulae, we can calculate the number of moles as shown below
$Number\,of\,moles\,=\,Molarity\left( M \right)\,\times \,Volume\,of\,solution\,$
Therefore, we can say that
Number of moles of nitric acid = $6.7\times 0.25=\,1.675$moles

Now we know that $number\,of\,moles\,=\,\dfrac{given\,mass\,of\,compound}{molar\,mass}$
In this question we need to find out the mass of the nitric acid. We can find the mass of the acid by modifying the above formula as
$mass\,of\,nitric\,acid\,(HN{{O}_{3}})=\,number\,of\,moles\,of\,acid\,\times \,molar\,mass\,of\,acid$
Now, firstly let us find out the molar mass of the nitric acid
We know that molar mass of hydrogen= 1g/mol
Molar mass of oxygen=$16 g/mol$
Molar mass of nitrogen=$14g/mol$
Therefore, Molar mass of $HN{{O} {3}}$= $1+\,14+48=63g/mol $
$\therefore mass\,of\,HN{{O}_{3}}=\,1.675\times \,63=\,105.5grams$ or $106 grams$
Hence, we can say that $106$ gram of nitric acid is present in the solution.

Note: We can use molarity for dilution of a solution as
${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$
Where ${{M}_{1\,}}and\,{{M}_{2}}\,$are initial and final molarity
${{V}_{1}}\,and\,{{V}_{2}}$are initial and final volume respectively.
Always remember to convert the volume into liters while solving the question related to the molarity. Always express the answer as per the rules of significant numbers and approximation for better calculations.