Question

Two samples of HCl of 1.0 M and 0.25 M are mixed. Find volumes of these samples taken in order to prepare a 0.75 M HCl solution. Assume no water is added.
(I) 20 mL, 10 mL (II) 100 mL, 50 mL (III) 40 mL, 20 mL (IV) 50 mL, 25 mL

a.) I, II, IV
b.) I, II
c.) II, III, IV
d.) I, II, III, IV

Answer 1

We can find the answer by using the molarity equation. In general the molarity equation can be given by :
Molarity of reactants $\times$Volume of reactants = Molarity of products $\times$Volume of products
Thus, (${M_1}{V_1}$ + ${M_2}{V_2}$) = ${M_3}$(${V_1} + {V_2}$)

Where ${M_1}$ is the molarity of the first solution.
${M_2}$ is the molarity of the second solution.
${M_3}$ is the molarity of the third solution.
And ${V_1},{V_2},{V_3}$ be the volumes of three solutions respectively.

Complete step by step answer:
First, let's write the things given to us and what we have to find.
We are given : Two solutions of HCl with different molarity.
Let the first solution be ${M_1}$.
And the second solution is ${M_2}$.
Molarity of first solution, ${M_1}$= 1.0 M
Molarity of second solution, ${M_2}$= 0.25 M
We are also given that on mixing these two solutions we get a third solution ${M_3}$ whose molarity = 0.75 M
We need to find : The volume of first two solutions to be mixed to form the solution of third molarity.
We know from the molarity equation that
Molarity of reactants $\times$Volume of reactants = Molarity of products $\times$Volume of products
Thus, (${M_1}{V_1}$+${M_2}{V_2}$) = ${M_3}$(${V_1} + {V_2}$)

So, on putting all the values, we have
(1.0 $\times {V_1}$+0.25$\times {V_2}$) = 0.75 (${V_1} + {V_2}$)
${V_1}$+ 0.25${V_2}$ = 0.75${V_1}$ + 0.75${V_2}$

On rearranging the equation, we have
${V_1}$ - 0.75${V_1}$ = 0.75${V_2}$- 0.25${V_2}$
0.25 ${V_1}$ = 0.5 ${V_2}$
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{0.5}}{{0.25}}$
It can be written as :
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{2}{1}$
So, the correct answer will be one in which the first solution will be double of the second solution.
Now, we will see the options one by one.

In (I) ${V_1}$= 20 mL and ${V_2}$= 10 mL
Here, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{2}{1}$. So, it is the correct answer.

In (II) ${V_1}$= 100 mL and ${V_2}$= 50 mL
Here also, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{2}{1}$. So, it is the correct answer.

In (III) ${V_1}$= 40 mL and ${V_2}$= 20 mL
Even in this, we see $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{2}{1}$. So, it is the correct answer.

In (IV) ${V_1}$= 50 mL and ${V_2}$= 25 mL
Even here, we see $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{2}{1}$. So, it is the correct answer.
So, the correct answer is “Option D”.

Note: The molarity of two solutions can never be added directly. For example - The equation can not be written as -
$({M_1} + {M_2}) = {M_3}$
It has to be - $({M_1}{V_1} + {M_2}{V_2}) = {M_3}{V_3}$