Question

The volumes of two $HCl$ solution A $\left( {0.5N} \right)$ and B $\left( {0.1N} \right)$ to be mixed for preparing $2L$ of $o.2N$ $HCl$ are:
A: $0.5L$ of A $ + $ $1.5L$ of B
B: $1.5L$ of A $ + $ $0.5L$ of B
C: $1L$ of A $ + $ $1L$ of B
D: $0.75L$ of A $ + $ $1.25L$ of B

Answer 1

Hint: Whenever two solutions are mixed together to form a solution of given normality or molarity then the sum of product of normality and volume of given solutions is equal to the product of normality and volume of the final solution.
Formula used: ${N_1}{V_1} + {N_2}{V_2} = NV$
${V_1} + {V_2} = V$
Where ${N_1},{V_1}$ is normality and volume of solution A, ${N_2},{V_2}$ is the normality and volume of solution B and $N,V$ is the normality and volume of final solution.

Complete step by step answer:
In this question volumes of two $HCl$ solution A $\left( {0.5N} \right)$ and B $\left( {0.1N} \right)$ are mixed to prepare $2L$ of $o.2N$ $HCl$ and we have to find the volume of solution A and solution B that should be mixed to form final solution. Using the formula ${N_1}{V_1} + {N_2}{V_2} = NV$ we can find relation between ${V_1}$ and ${V_2}$ and the second equation is ${V_1} + {V_2} = V$.
Substituting the given values in the equation ${N_1}{V_1} + {N_2}{V_2} = NV$:
$\left( {0.5 \times {V_1}} \right) + \left( {0.1 \times {V_2}} \right) = 0.2 \times 2$
$0.5{V_1} + 0.1{V_2} = 0.4$
This equation can be written as:
$5{V_1} + {V_2} = 4$ (Equation a)
Final volume of the solution is $2L$. Substituting this in the equation ${V_1} + {V_2} = V$,
${V_1} + {V_2} = 2$ (Equation b)
Solving the equation a and equation b we get:
${V_1} = 0.5L$ and ${V_2} = 1.5L$
This means volume of solution A is $0.5L$ and volume of solution B is $1.5L$.
So, correct answer is option A that is $0.5L$ of A $ + $ $1.5L$ of B.

Note:
Normality of a solution is defined as the number of gram equivalents present per liter of solution. Number of gram equivalents can be found by multiplying the number of moles of a substance with valence of that substance.