Question

0.5g of a mixture of ${{K}_{2}}C{{O}_{3}}$ and $L{{i}_{2}}C{{O}_{3}}$ requires 30 mL of 0.25N HCl solution for neutralization. The percentage composition of the mixture is:
(A)- 96% ${{K}_{2}}C{{O}_{3}}$, 4% $L{{i}_{2}}C{{O}_{3}}$
(B)- 4% ${{K}_{2}}C{{O}_{3}}$, 96% $L{{i}_{2}}C{{O}_{3}}$
(C)- 48% ${{K}_{2}}C{{O}_{3}}$, 52% $L{{i}_{2}}C{{O}_{3}}$
(D)- 52% ${{K}_{2}}C{{O}_{3}}$, 48% $L{{i}_{2}}C{{O}_{3}}$

Answer 1

Normality (N) is defined as the number of gram equivalents of a solute present per liter of the solution. It is given as:
$\text{Normality}=\dfrac{\text{weight of solute (in grams)}}{\text{equivalent weight}}\times \dfrac{1000}{\text{volume of solution (in mL)}}$.

Equivalent weight of a compound is given as:
$\text{Equivalent weight = }\dfrac{\text{Molar mass}}{\text{number of electrons lost or gained}}$

Complete step by step answer:
We have been given that the sum of amount of ${{K}_{2}}C{{O}_{3}}$ and $L{{i}_{2}}C{{O}_{3}}$ is equal to 0.5g. If the weight of ${{K}_{2}}C{{O}_{3}}$ in the mixture is taken to be ‘x’ g then, the amount of $L{{i}_{2}}C{{O}_{3}}$ in the mixture will be (0.5-x) g.

Let us now calculate the equivalent weights of ${{K}_{2}}C{{O}_{3}}$ and $L{{i}_{2}}C{{O}_{3}}$.
Molar mass of ${{K}_{2}}C{{O}_{3}}$ is calculated as: $39\times 2+12+3\times 16=138gmo{{l}^{-1}}$
Therefore, the equivalent weight of ${{K}_{2}}C{{O}_{3}}$ will be equal to

$\text{Equivalent weight of }{{\text{K}}_{2}}\text{C}{{\text{O}}_{3}}\text{ = }\dfrac{138gmo{{l}^{-1}}}{2}=69g\text{ }e{{q}^{-1}}$
Molar mass of $L{{i}_{2}}C{{O}_{3}}$ is calculated to be: $7\times 2+12+3\times 16=74g\,mo{{l}^{-1}}$
Equivalent weight of $L{{i}_{2}}C{{O}_{3}}$ can now be calculated by dividing its molar mass by 2.
$\text{Equivalent weight of L}{{\text{i}}_{2}}\text{C}{{\text{O}}_{3}}\text{ = }\dfrac{74gmo{{l}^{-1}}}{2}=37g\text{ }e{{q}^{-1}}$

According to the question, the mixture of ${{K}_{2}}C{{O}_{3}}$ and $L{{i}_{2}}C{{O}_{3}}$ is completely neutralized by 30mL of 0.25N HCl. So, we can write for complete neutralization reaction
n equivalent of ${{K}_{2}}C{{O}_{3}}$ + n equivalent of $L{{i}_{2}}C{{O}_{3}}$ = n equivalent of HCl

Now, let us find the number of equivalents of HCl.
We know that normality is given as:
$\text{Normality}=\dfrac{\text{weight of solute (in grams)}}{\text{equivalent weight}}\times \dfrac{1000}{\text{volume of solution (in mL)}}$
And number of equivalent, n is equal to
$n=\dfrac{\text{weight of solute (in grams)}}{\text{equivalent weight}}$
So the above equation now becomes
$\text{Normality}=n\times \dfrac{1000}{\text{volume of solution (in mL)}}$
Or, we can rewrite it as
$n=\text{Normality}\times \dfrac{\text{volume of solution (in mL)}}{1000}$

Given, normality of HCl solution, N = 0.25 N
Volume of HCl required to neutralize ${{K}_{2}}C{{O}_{3}}$ and $L{{i}_{2}}C{{O}_{3}}$, V = 30 mL
Substituting the values of volume and normality of HCl, we can fine n equivalent of HCl as follows:
$n=\text{Normality}\times \dfrac{\text{volume of solution (in mL)}}{1000}$
$n=0.25N\times \dfrac{30mL}{1000}=0.0075$

Number of equivalents of ${{K}_{2}}C{{O}_{3}}$ is calculated as:
$n=\dfrac{\text{weight of }{{K}_{2}}C{{O}_{3}}}{\text{equivalent weight of }{{K}_{2}}C{{O}_{3}}}=\dfrac{x}{69}$
Similarly, number of equivalents of $L{{i}_{2}}C{{O}_{3}}$ is calculated as:
$n=\dfrac{\text{weight of }L{{i}_{2}}C{{O}_{3}}}{\text{equivalent weight of }L{{i}_{2}}C{{O}_{3}}}$
$n=\dfrac{0.5-x}{37}$

Now, putting the values of n in the equation given below, we obtain
n equivalent of ${{K}_{2}}C{{O}_{3}}$ + n equivalent of $L{{i}_{2}}C{{O}_{3}}$ = n equivalent of HCl
$\dfrac{x}{69}+\dfrac{0.5-x}{37}=0.0075$
$\dfrac{37x+69(0.5-x)}{69\times 37}=0.0075$
$37x+34.5-69x=19.1475$
$34.5-32x=19.1475$
Solving the above equation for x, we get
$15.3525=32x$
$x=\dfrac{15.3525}{32}=0.48$
The weight of ${{K}_{2}}C{{O}_{3}}$ in the mixture, x = 0.48g.
Thus, the amount of $L{{i}_{2}}C{{O}_{3}}$, (0.5-x) g will be (0.5-0.48) g = 0.2g.

Therefore, the percentage of ${{K}_{2}}C{{O}_{3}}$ in the mixture = $\dfrac{\text{weight of }{{\text{K}}_{2}}\text{C}{{\text{O}}_{3}}(x)}{0.5g}\times 100$
Substituting the value of x in the above equation, we get = $\dfrac{\text{weight of }{{\text{K}}_{2}}\text{C}{{\text{O}}_{3}}(x)}{0.5g}\times 100$
Percentage composition of $L{{i}_{2}}C{{O}_{3}}$ will be (100-96) % = 4%.
So, the correct answer is “Option A”.

Note: Solve this question step by step to avoid errors. Carefully perform all the calculations. Do not get confused between the molar mass, equivalent weights and number of equivalents.