Question

If 100ml of 1M ${{H}_{2}}S{{O}_{4}}$ is mixed with 100ml of 98% $\left( \dfrac{w}{w} \right)$ of ${{H}_{2}}S{{O}_{4}}$solution (d = 0.1g/L) then,
[A] Concentration of solution becomes half.
[B] Volume of solution becomes 200ml.
[C] Mass of sulphuric acid in the solution is 98g.
[D] Mass of sulphuric acid in the solution is 19.6g.

Answer 1

Hint: This question has multiple correct answers therefore, to find which of the given options are correct let us find out the concentration of the solution, mass of sulphuric acid in the solution and the volume of the solution. Comparing the calculated values with the given option will help us identify which of the given options is correct.

Complete answer:
Firstly, let us calculate the concentration of the solution. We know that the concentration of a solution is the number of solutes dissolved in the solvent. The concentration of the first solution is given to us which is 1M and to this 100ml 98% sulphuric acid was added.
For the second solution, the molarity will be-
     \[M=\dfrac{wt.%\times density\times 10}{molecular\text{ weight}}\]
We know that the molecular weight of sulphuric acid is 98 and the weight percentage and the density is given to us. Therefore, putting these values in the equation we will get-
     \[M=\dfrac{98\times 0.1\times 10}{98}=1M\]
As we can see that the concentration of the first solution was 1M and the second solution which was added to it is 1M too.
Now let us check the molarity of the mixture of the two solutions.
The total number of moles = ${{V}_{1}}{{M}_{1}}+{{V}_{2}}{{M}_{2}}$, where V is the volume and M is the molar concentration of the first and the second solution. The number of moles will be in ‘millimoles’.
So, total number of moles = $\left( 100ml\times 1M \right)+\left( 100ml\times 1M \right)\text{ =200Mmol}$
Therefore, molarity = $\dfrac{no.\text{ }of\text{ moles in Mmol}}{volume\text{ in mL}}=\dfrac{200}{200}=1M$

The concentration of the new solution is also 1M. Therefore, there is no change in the concentration of the solution. Therefore, the first option is incorrect.
Next, let us check the volume. The first solution was 100ml and the second was 100ml too. So, the total volume of the solution is 200ml. Therefore, the second option is correct.
Now for the mass of sulphuric acid-
Number of moles in 1L or 1000mL of the solution = 1 (as the molarity is 1M) = 98g (molecular mass)
So, mass of sulphuric acid in 1000mL = 98g
Therefore, mass in 200mL = $\dfrac{98}{1000}\times 200=19.6g$

Therefore, the correct answers are option [B] Volume of solution becomes 200ml and [D] Mass of sulphuric acid in the solution is 19.6g.

Note: Here 200M mol is equal to 0.2mol. If we used this value i.e. in mol, we would have to convert the 200mL of solution to L. Then, the molarity of the solution would be, $molarity=\dfrac{0.2}{0.2}=1M$, which is the same as we calculated above.
Here, we can also calculate the number of moles in the mixture of solution by using the volume. If we multiply the volume by the density, we will get the mass and dividing the mass by the molar mass will give us the number of moles.