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

Hint: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.
\[{\text{ppm(A)}} = \dfrac{{{\text{Mass of A}}}}{{{\text{Total mass of the solution}}}} \times {10^6}\]