Question

Volume of water required to convert 100ml of $0.1{\text{M}}\,{K_2}C{r_2}{O_7}/{H^ + }\;$ to $0.1{\text{N}}\;$ is:
A.$100{\text{ml}}$
B.$500{\text{ml}}$
C.$400{\text{ml}}$
D.$200{\text{ml}}$

Answer 1

To answer this question, you should recall the concept of normality. Normality: It is defined as the number of gram equivalents of solute present in one litre of the solution.
The formula used:
${\text{normality = molarity}} \times {\text{valence factor}}$

Complete step by step answer:
On dilution, the moles ${K_2}C{r_{2}}{O_7}$ will remain constant.
Now, ${K_2}C{r_{2}}{O_7}$, we can write the reaction in the acidic medium as
${K_2}C{r_2}{O_7}{\text{ + }}14{H^ + } + 6{e^ - } \to 2{K^ + } + 2C{r^{3 + }}{\text{ + }}7{H_2}O$
Here we can see the n-factor of ${K_2}C{r_{2}}{O_7}$ in acidic medium ​ is 6.
Let the required volume be $x\,{\text{ml}}$.
We will be able to write the normality of this solution as:
$\Rightarrow \left( {\dfrac{{0.1 \times 100}}{{100 + x}}} \right) \times 6$
Now equating this value to the normality given in the question:
$\Rightarrow \left( {\dfrac{{0.1 \times 100}}{{100 + x}}} \right) \times 6 = 0.1$
Solving this we will get the required volume as: $500{\text{ml}}$.

Hence, the correct answer to this question is option B.

Note:
Other concentration terms commonly used are:
Concentration in Parts Per Million (ppm) The parts of a component per million parts (${10^6}$) of the solution.
${\text{ppm(A)}} = \dfrac{{{\text{Mass of A}}}}{{{\text{Total mass of the solution}}}} \times {10^6}$