Question

I. If ${ 5.5g }$ of a mixture of ${ Fe }{ SO }_{ 4 }{ .7H }_{ 2 }{ O }$ and ${ Fe }_{ 2 }\left( { SO }_{ 4 } \right) _{ 3 }{ .9H }_{ 2 }{ O }$ required ${ 5.4mL}$ of 0.1N ${ Fe }{ SO }_{ 4 }{ .7H }_{ 2 }{ O }$ solution for II. complete oxidation, then calculate the number of gram moles of hydrated ferric sulphate in mixture.
The vapour density of a mixture consisting of ${ NO }_{ 2 }$ and ${ N }_{ 2 }{ O }_{ 4 }$ is 38.3 at ${ 26.7 }^{ \circ }{ C }$. Calculate the number of moles of ${ NO }_{ 2 }$ in ${ 100g }$ of the mixture.
A. (i) ${ 4.76mmol }$, (ii) ${ 0.265mol }$
B. (i) ${ 5.36mmol }$, (ii) ${ 0.34mol }$
C. (i) ${ 9.52mmol }$, (ii) ${ 0.43mol }$
D. (i) ${ 10.72mmol} $, (ii) ${ 0.68mol }$

Answer 1

Hint: Vapour density: It is defined as the ratio of the mass of a volume of a gas, to the mass of an equal volume of hydrogen, measured under the same conditions of temperature and pressure.
\[Vapour\text{ }density=\dfrac{molar\text{ mass}}{2}\]

Complete Solution :
Let x be the number of gram moles of hydrated ferric sulphate in mixture.
It is given that,
Total mass of mixture = ${ 5.5g }$
The molar mass of hydrated ferric sulphate ${ Fe }_{ 2 }\left( { SO }_{ 4 } \right) _{ 3 }{ .9H }_{ 2 }{ O }$ is ${ 562gmol }^{ -1 }$.
The mass of hydrated ferric sulphate will be ${ 562x }$ grams.
So, the mass of hydrated ferrous sulphate will be ${ 5.5-562x grams }$.
The molar mass of hydrated ferrous sulphate is ${ 278 gmol }^{ -1 }$.
The number of moles of hydrated ferrous sulphate will be
\[\dfrac{5.5-562x}{278}\]
This is equal to the number of moles of ferrous ions that can be oxidized by potassium permanganate.
5.4mL of 0.1N potassium permanganate are used.

Hence, the number of gram equivalent of potassium permanganate:
\[0.1\times \dfrac{5.4}{1000}=0.00054eq\]
They are also equal to the number of g eq. of ferrous ions. They are also equal to the number of moles of ferrous ions.
$\dfrac { 5.5-562x }{ 278 } { =0.00054 }$
$5.5-562x=0.15012$
${ 562x=5.349 }$
${ x= }\dfrac { 5.349 }{ 562 } $
${ x=9.52\times 10 }^{ -3 }{ moles }$
x = ${ 9.52mmol }$
Hence, the number of gram moles of hydrated ferric sulphate in mixture is ${ 9.52 mmol }$.

(ii) Let ${ 100g }$ of mixture contains x moles of ${ NO }_{ 2 }$
The molecular weight of ${ NO }_{ 2 }$ is ${ 46 gmol }^{ -1 }$.
The mass of x moles will be 46x grams

So, the mixture will contain ${ 100-46x }$ grams of ${ N }_{ 2 }{ O }_{ 4 }$.
The molecular weight of ${ N }_{ 2 }{ O }_{ 4 }$ is ${ 92 gmol }^{ -1 }$.
Hence, the number of moles of ${ N }_{ 2 }{ O }_{ 4 }$;
$\dfrac { 100-46x }{ 92 gmol^{ -1 } } $

Therefore, the average molecular mass of the mixture is:
$\dfrac { 46x+92\left( \dfrac { 100-46x }{ 92 } \right) }{ x+\left( \dfrac { 100-46x }{ 92 } \right) } $
$\dfrac { 100 }{ x+\left( \dfrac { 100-46x }{ 92 } \right) } $
$\dfrac { 9200 }{ 92x+100-46x } $
$\dfrac { 9200 }{ 46x+100 } $
It is given that,
The vapour density of the mixture = ${ 38.3 }$.

As we know, Molecular weight = ${ 2\times 38.3=76.6 gmol^{ -1 } }$
= ${ 2\times 38.3=76.6 gmol^{ -1 } }$…..(2)
From equation (1) and (2), we get
$\dfrac { 9200 }{ 46x+100 } { =76.6 }$
${ 120.1 = 46x+100 }$
${ 46x = 20.1 }$
${ x= }\dfrac { 20.1 }{ 46 } $
x = ${ 0.437 mol }$
Hence, the number of moles of ${ NO }_{ 2 }$ in ${ 100 g }$ of the mixture = ${ 0.437 mol }$
So, the correct answer is “Option C”.

Note: The possibility to make a mistake is that to calculate the gram moles of hydrated ferric sulphate, you must have to calculate the number of gram equivalent of potassium permanganate. Also, the molecular mass is equal to the vapour density multiplied by 2.