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Question: The respective volumes (in ml) of \(0.2M\text{ HCl}\)and \(0.6M\text{ HCl}\) to be mixed together fo...

The respective volumes (in ml) of 0.2M HCl0.2M\text{ HCl}and 0.6M HCl0.6M\text{ HCl} to be mixed together for preparing 1000 ml of0.3M HCl1000\text{ }ml\text{ }of 0.3M\text{ }HCl are :
(a) 300 and 700 300\text{ and 700 }
(b)750 and 250 750\text{ and 250 }
(c) 250 and 750 250\text{ and 750 }
(d) 500 and 500 500\text{ and 500 }

Explanation

Solution

We have been given the molarities of the solutions of HClHCl which combines together to form the 1000 ml of0.3M HCl1000\text{ }ml\text{ }of0.3M\text{ }HCl. So, thus we can easily find the volume of the solutions by assuming them as xx and 1000−x1000-x as total volume of solution is 1000ml1000ml and applying the formula as: M1V1+M2V2=M3V3{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}={{M}_{3}}{{V}_{3}}. Now you can easily solve it.

Complete Solution :
First of let’s discuss what is molarity. By the term molarity we mean, the no. of moles of the solute to the volume of the solution in litres or milliliters. It denoted as M. So,
molarity(M)=no. of moles of the solutevolume of the solution in litersmolarity(M)=\dfrac{no.\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{volume\text{ }of\text{ }the\text{ }solution\text{ }in\text{ }liters}

- Now considering the statement:
We can find the volume of the solutions having the molarities as 0.2M0.2M and 0.6M0.6M which are mixed together to prepare 1000 ml of0.3M HCl1000\text{ }ml\text{ } of 0.3M\text{ }HCl as;
M1V1+M2V2=M3V3{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}={{M}_{3}}{{V}_{3}} ----------------(1)
As we know that:
molarity, M1=0.2M{{M}_{1}}=0.2M(given)
molarity, M2=0.6M{{M}_{2}}=0.6M(given)
molarity, M3=0.3M{{M}_{3}}=0.3M(given)
V3=1000 ml{{V}_{3}}=1000\text{ ml}

So. let’s suppose that:
V1=x and V2=1000−x \begin{aligned} & {{V}_{1}}=x\text{ and} \\\ & {{V}_{2}}=1000-x \\\ \end{aligned}

Now, put these all values in equation (1), we get:
0.2×x+0.6×(1000−x)=0.3×1000 0.2x+600−0.6x=300 0.4x=600−300 x=3000.4  =750 \begin{aligned} & 0.2\times x+0.6\times (1000-x)=0.3\times 1000 \\\ & 0.2x+600-0.6x=300 \\\ & 0.4x=600-300 \\\ & x=\dfrac{300}{0.4} \\\ & \text{ =750} \\\ \end{aligned}

So, the value of :
V1=750 ml and V2=1000−750 ml  = 250 ml \begin{aligned} & {{V}_{1}}=750\text{ }ml\text{ and} \\\ & {{V}_{2}}=1000-750\text{ }ml \\\ & \text{ = 250 }ml \\\ \end{aligned}
Hence, the volumes of the solutions having the molarities as 0.2M0.2M and 0.6M0.6M which are mixed together to prepare 1000 ml of0.3M HCl1000\text{ }ml\text{ }of0.3M\text{ }HCl are 750 ml and 250 ml750\text{ ml and 250 ml}.
So, the correct answer is “Option B”.

Note: Molarity of the solution changes with change in the temperature of the solution because with the increase in the temperature , the volume of the solution changes and hence the molarity changes. If temperature is increased, volume is increased and hence molarity decreases. On the contrary, if temperature is decreased , volume is decreased but the molarity increases. So, this means that the molarity and the volume are inversely proportional to each other.