Question

The mass of \[{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\] required to produce \[5.0{\text{ L C}}{{\text{O}}_2}\] at \[{77^ \circ }{\text{C}}\] and \[0.82{\text{ atm}}\]pressure from excess of oxalic acid and volume of \[0.1{\text{ N NaOH}}\] required to neutralize the \[{\text{C}}{{\text{O}}_2}\] evolved respectively, are:
A.\[7{\text{ g, }}2.86{\text{ L}}\]
B.\[{\text{5 g, 1}}.86{\text{ L}}\]
C.\[{\text{4 g, 0}}.86{\text{ L}}\]
D.None of the above

Answer 1

Hint: The above question is to be done in two steps, in the first step we have to calculate the number of moles of carbon dioxide using the ideal gas equation, all should be in Standard units and hence, to find out mass of \[{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\] using the reaction equation. Carbon dioxide is present as carbonic acid in aqueous solution. For the neutralization number equivalent of \[{\text{C}}{{\text{O}}_2}\] and sodium hydroxide is the same.
Formula used:
\[{\text{PV}} = {\text{nRT}}\] where P is pressure, V is volume of gas, n is the number of moles, R is gas constant and T is temperature.
\[{\text{number of moles}} = \dfrac{{{\text{mass}}}}{{{\text{molecular mass}}}}\]

Complete step by step answer:
Let us first calculate the number of moles of carbon dioxide using ideal gas equation:
We will take the value of \[{\text{R}} = 0.082{\text{ l atm }}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}\],To convert temperature into Kelvin we need we need to ass 273 to the temperature in degree Celsius.
\[T({\text{in K}}) = {77^ \circ }{\text{C}} + 273 = 350{\text{K}}\]
Substituting the given values we will get:
\[\dfrac{{{\text{PV}}}}{{{\text{RT}}}} = {\text{n}}\]
\[ \Rightarrow \dfrac{{{\text{0}}{\text{.82}} \times {\text{5}}}}{{{\text{0}}{\text{.082}} \times 350}} = {\text{n}}\]
Solving this:
\[{\text{n}} = \dfrac{1}{7}\]
The reaction of \[{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\] with oxalic acid is:
\[{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7} + 7{{\text{H}}_2}{{\text{C}}_2}{{\text{O}}_4} + 2{{\text{K}}_2}{{\text{C}}_2}{{\text{O}}_4} \to 2{{\text{K}}_3}[{\text{Cr}}{({{\text{C}}_2}{{\text{O}}_4})_3}] + 6{\text{C}}{{\text{O}}_2} + 7{{\text{H}}_2}{\text{O}}\]
For 6 moles of \[{\text{C}}{{\text{O}}_2}\] 1 mole of \[{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\] is required hence for I mole of \[{\text{C}}{{\text{O}}_2}\], \[\dfrac{1}{6}\] mole of \[{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\] is required. Similarly for \[\dfrac{1}{7}\] moles of \[{\text{C}}{{\text{O}}_2}\] , \[\dfrac{1}{{7 \times 6}}\] mole of \[{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\] is required.
Now we will calculate the mass of \[{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}\] form the above formula:
\[{\text{mass }} = \dfrac{{\text{1}}}{{7 \times 6}} \times 294 = 7{\text{ g}}\]
The n- factor for carbonic acid that is \[{H_2}C{O_3}\] is 2. So we have to multiply mole of carbon dioxide with this to get gram equivalent:
No. of equivalent for \[{H_2}C{O_3}\] is \[\dfrac{2}{7}\].
And the number of equivalent sodium hydroxide is \[0.1{\text{ N}} \times {\text{V}}\]. For neutralization both should be equal:
\[{\text{V}} = 2.86{\text{L}}\]

The correct option is A.

Note:
Equivalent weight as the name suggests is the mass of 1 equivalent. It is also known as the gram equivalent they both are the same thing. It is a mass of a given substance that combines with fixed mass of a fixed quantity of another substance.