Question

Two sulphuric acid solutions are mixed as follows: \[\]$25 \cdot 0mt$of a $0.50M$ Sulphuric acid solution are added to $0 \cdot 075L$ of a $0 \cdot 25M$ sulphuric acid solution. What is the molarity of the muhing mixture? (Assume volumes are additive)
A. $0 \cdot 501M$
B. $0 \cdot 312M$
C. $0 \cdot 712M$
D. $0 \cdot 12M$

Answer 1

Hint: The sulphuric acid which has the chemical formula as ${H_2}S{O_4}$ is also known as oil of vitriol. It is a mineral acid that is composed of sulphur, oxygen and hydrogen elements in it. It is colourless, odourless and viscous liquid that is soluble in water.

Formula used:
Concentration of final sulphuric acid mixture, the molarity formula,
$ = \dfrac{{{M_1}{V_1} + {M_2}{V_2}}}{{\text{Total Volume}}}$

Complete answer:
Here in the question that is given down, as we focused that is it the direct substitution into the formula, we have to look in to the values before substituting,
${M_1} = 0 \cdot 50M$
${V_1} = \dfrac{{25}}{{1000}} = 0 \cdot 025L$
Here, ${M_1}$ and ${V_1}$ are the molarity and volume of the first sulphuric acid. Here the volume is directly converted from mili-litre to litre.
${M_2} = 0 \cdot 25M$
${V_2} = 0 \cdot 075L$
And now in order to obtain the total volume we have to add up the two volumes of the two sulphuric acid.
$V = 0 \cdot 025 + 0 \cdot 075 = 0 \cdot 1L$
Now coming to the main part of the question, we have our final concentration,
$ = \dfrac{{{M_1}{V_1} + {M_2}{V_2}}}{{\text{Total Volume}}}$
Now it is all the direct substitution,
$ = \dfrac{{\left( {0 \cdot 5 \times 0 \cdot 025} \right) + \left( {0 \cdot 25 \times 0 \cdot 075} \right)}}{{0 \cdot 1}}$
After the calculation we get our answer which then we have to look into the options to see if there is the actual solution to look into the nearer answer to them.
$ = 0 \cdot 3125M$
After looking into the final answer we know the right option already,

Hence, the right option is (B) .

Note:
Molarity changes with the temperature because the volume changes with the temperature. So, we need to measure molarity by keeping temperature unchanged. So , in a lab when you are preparing the solution make sure to note down the temperature otherwise it will affect the calculations