Question

When V ml of 2.2 M ${H_2}S{O_4}$ solution is mixed with 10 V ml of water, the volume contraction of 2% takes place. The molarity of diluted solution is:
A. 0.204 M
B. 0.265 M
C. 0.345 M
D. 1.002 M

Answer 1

To solve this question one must know about molarity, molarity is the ratio of moles of solute present in 1 litre of solution. It is represented by M. The formula of molarity is:
$M = \dfrac{n}{v}$, where M = molarity of the solution, n = number of moles of the solute, and v = volume of the solution.

Complete answer:
As we all know, molarity is generally expressed in units of moles of solute per litre of solution. We also know that the formula for molarity is given by $M = \dfrac{n}{v}$, where M = molarity of the solution, n = number of moles of the solute, and v = volume of the solution.
Let, ${M_1}$ be the molarity of the ${H_2}S{O_4}$ solution and ${V_1}$ be its volume. It is given in the question that, ${M_1=}$ 2.2 M and ${V_1=}$ V ml.
Let the molarity of the diluted solution be ${M_2}$ and the final volume of the of the solution after adding 10V ml water be ${V_2} = 10V + {V_1}$ ml
We have to calculate the value of ${M_2}$.
Now we know that, ${M_1}{V_1} = {M_2}{V_2}$
Now, the total volume of final solution =${V_2} = 10V + {V_1}$ = 11V ml
It is also given in the question that the volume of the final solution has contracted by 2%.
The amount of volume contracted can be calculated by using this formula:
Contacted volume = [(Original Volume)-(% of contraction)${\times}$(Original Volume)]
Therefore, Contacted volume = $\left[ {11V - \dfrac{2}{{100}} \times 11V} \right] = 0.98 \times 11 \quad Vml$
Now using, ${M_1}{V_1} = {M_2}{V_2}$
$\Rightarrow$ $2.2 \times V = {M_2}(0.98 \times 11)$
$\Rightarrow$ ${M_2} = 0.204M$
So the molarity of diluted solution is 0.204 M.
So, the correct answer is “Option A”.

Note: As we have seen in this solution that molarity depends on the volume of the solution. We know that with an increase in temperature, the volume also increases. Thus molarity depends on the temperature.
In order to remove the temperature dependency of the concentration term we can convert the molarity into molality. Molality is the ratio of the moles of solute present in 1 kg of the solvent. Since molality depends on the mass and not on the volume, its value won’t be affected by the temperature.