Question

What volume of water should be added to 50ml of \[HN{{O}_{3}}\] having density d = 1.5 and $63\%$ by weight to have 1M solution.
A.750 ml
B.700ml
C.800 ml
D.1400 ml

Answer 1

Hint: The question requires an answer of the volume of water that should be added in 50 ml of \[HN{{O}_{3}}\] , that means the total volume will be more than the volume of water added. Assumption of weight of solution is needed to answer this question.

Complete answer:
The correct answer to this question is option B, 700 ml water should be added to 50 ml \[HN{{O}_{3}}\] to have a 1M solution.
Let us see how, first as the weight percent of \[HN{{O}_{3}}\] (nitric acid) is 63, thus let us take 100 gm of solution for reference.
Therefore, 63 gram of \[HN{{O}_{3}}\] is present in 100 gram of solution. Now, let’s calculate number of moles first, therefore
Number of moles = \[\dfrac{G.M.}{M.M.}\]
Here, G.M. represents given mass and M.M. represents the molar mass of the compound.
Thus, the molar mass of \[HN{{O}_{3}}\] is 63 gram per mole.
Number of moles = \[\dfrac{63g}{63\dfrac{g}{mol}}\]
Therefore, Number of moles = \[1mol\]
Now, as density and mass are given, we can find out the volume of the solution.
\[Density=\dfrac{Mass}{Volume}\]
Therefore, \[Volume=\dfrac{Mass}{Density}\]
Putting the values of mass and density in the above equation.
\[Volume=\dfrac{100g}{1.5\dfrac{g}{ml}}\]
\[Volume=66.7ml\]
The mass has been taken as 100 grams because we assumed the mass of the solution as 100 grams.
Now, let us find the molarity of the solution with the volume of 66.7 ml.
Molarity is the number of moles of solute per litre of solution, its formula is given below.
\[Molarity=\dfrac{n\times 1000}{V}\]
Here, 1000 has been multiplied in the numerator as the volume is in ml and not in litres.
Putting the values of n and V in the above given reaction.
\[Molarity=\dfrac{1\times 1000}{66.7}\]
\[Molarity=15M\]
Therefore, the molarity of the solution with volume of 66.7 ml is 15M, now to find the volume of solution with molarity of 1M let us use the formula of neutralization.
\[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\]
We have found the molarity \[{{M}_{1}}\] , the volume \[{{V}_{1}}\] and we have been given in the question that the molarity \[{{M}_{2}}\] is given as 1M. Thus, putting all the given values in the above given equation.
\[15\times 50=1\times {{V}_{2}}\]
Therefore, \[{{V}_{2}}=750ml\]
Therefore, the volume of the solution for molarity to be 1M should be 750ml.
Now, the solution contains 50 ml of \[HN{{O}_{3}}\] (nitric acid), thus the volume of water in the solution should be given by
Volume of water = Volume of Solution – Volume of \[HN{{O}_{3}}\]
Therefore, Volume of water = 750 – 50 = 700 ml
Therefore, the volume of water in the solution having molarity of 1M, should be 700 ml.
And hence the correct answer is option B.

Note:
Care should be taken while reading this type of questions, the options also indicate 750 ml as one solution and 700 ml as one solution. Therefore, one might get confused about the final answer to the question. The molar mass of \[HN{{O}_{3}}\] (nitric acid) is 63, but doing simple calculations one can get the answer as 64, thus the final answer may vary a little bit.