Question

Volume of water required to convert 100 ml of $ 0.1{\text{M }}{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}/{{\text{H}}^ + } $ to 0.1 N is:
(A) 100 ml
(B) 200 ml
(C) 500 ml
(D) 400 ml

Answer 1

Hint: Concentration is defined as the abundance of an ingredient divided by the total volume of a mixture in chemistry. Mass concentration, molar concentration, number concentration, and volume concentration are all examples of mathematical descriptions. A concentration can be any chemical combination, though solutes and solvents in solutions are the most common. Normal concentration and osmotic concentration are two types of molar (amount) concentration.

Complete answer:
Normality is defined as the number of gram or mole equivalents of solute contained in one litre of a solution, according to the standard definition. The amount of moles of reactive units in a chemical is what we mean when we say equivalent. The number of mole equivalents per litre of solution is defined as normality (N).
 $ {\text{Normality = }}\dfrac{{{\text{Number of mole equivalents}}}}{{{\text{1 L of solution}}}} $
Normality, like molarity, describes the relationship between the quantity of solute in a solution and the total volume of the solution; however, normality is only applied to acids and bases.
The quantity of a material in a given volume of solution is measured in molarity (M). The moles of a solute per litre of a solution is known as molarity. The molar concentration of a solution is also known as molarity.
Now coming to the question, let’s say the needed capacity is x ml. Now, normality equals molarity multiplied by the valence factor.
Valence factor of $ {{\mathbf{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}/{{\text{H}}^ + } = 6 $
Now hence the resulting molarity will be $ = \dfrac{{0.1 \times 100}}{{100 + x}} = \dfrac{{10}}{{100 + x}} $
Resulting molarity = $ = \dfrac{{10}}{{100 + x}} \times 6 $
Hence,
 $ 0.1 = \dfrac{{60}}{{(100 + x)}} $
 $ \Rightarrow 0.1{\text{x}} = 50 $
 $ \Rightarrow {\mathbf{x}} = 500{\text{ml}} $
Hence option C is correct.

Note:
The use of molar concentration in thermodynamics is frequently inconvenient since the volume of most solutions varies somewhat with temperature owing to thermal expansion. This difficulty is typically overcome by utilising temperature adjustment factors or a temperature-independent concentration measure such as molality.