Tag: gas laws and mole concept

Questions Related to gas laws and mole concept

A definite mass of $ H _{2}O _{2} $ is oxidized by excess of acidified $ KMnO _{4} $ and acidified $ K _{2}Cr _{2}O _{7} $, in separate experiments. Which of the following is/are correct statements? 
(K = 39, Cr = 52, Mn = 55 )

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. Mass of $ K _{2}Cr _{2}O _{7} $ used up will be greater than that of $ KMnO _{4} $

  2. Moles of $ KMnO _{4} $ used up will be greater than that of $ K _{2}Cr _{2}O _{7} $

  3. Equal mass of oxygen gas is evolved in both the experiments.

  4. If equal volumes of both the solutions are used for complete reaction, then the molarities of $ KMnO _{4} $ and $ K _{2}Cr _{2}O _{7} $ solutions are in $6:5$ ratio.

Show answer
Correct Option: A,B,D
Explanation:

According to question, reaction of experiment (1) and (2)
(1)$ 5H _{2}O _2+2KMnO _{4}+3H _{2}SO _{4}\rightarrow 5O _{2}+2MnSO _{4}+K _{2}SO _{4}+8H _{2}O $


(2)$ 3H _{2}O _{2}+K _{2}Cr _{2}O _{7}+4H _2SO _{4}\rightarrow Cr _{2}(SO _{4}) _{3}+3O _{2}+K _{2}SO _{4}+7H _{2}O $

(a) According to reaction (1)
5 mole $ H _{2}O _{2} = 2\,mole KMnO _{4} $

1 mole $ H _{2}O _{2} = \dfrac{2}{5} = 0.4\,mole\,KMnO _{4} = 63.2\,g\,KMnO _{4} $

According to reaction (2)
3 mole $ H _{2}O _{2} = 1\,mole\,K _{2}Cr _{2}O _{7} $

1 mole $ H _{2}O _{2} = \dfrac{1}{3} mole\,K _{2}Cr _{2}O _{7} = 98.1 g \, K _{2}Cr _{2}O _{7} $

Mass of $ K _{2}Cr _{2}O _{7}> KMnO _{4} $

(b) Moles of $ KMnO _{4}> $ moles of $ K _{2}Cr _{2}O _{7} (0.333) $

(c)In reaction 1, 5 Mole $ H _{2}O _{2} $ released = 5 mole $ O _{2}\Rightarrow 32\times 5=160g $
In reaction 2, 3 mole $ H _{2}O _{2} $ released = 3 mole $ O _{2}=3\times 32= 96 g$ 

(d) 1 mole of $ H _{2}O _{2} = \dfrac{2}{5} $ moles $ KMnO _{4} $ exp...(1)
1 mole of $ H _{2}O _{2} = \dfrac{1}{3}$ mole $ K _{2}Cr _{2}O _{7} $ exp...(2)
$ \dfrac{KMnO _{4}}{K _{2}Cr _{2}O _{7}} = \dfrac{\dfrac{2}{5}}{\dfrac{1}{3}} = \dfrac{2}{5}\times \dfrac{3}{1} = \dfrac{6}{5}\Rightarrow 6:5 $

Options A, B and D are correct.

Two acids $ H _{2}SO _{4} $ and $ H _{3}PO _{4} $ are neutralized separately by the same amount of an alkali when sulphate and dihydrogen orthophosphate are formed, respectively. Find the ratio of the masses of $ H _{2}SO _{4} $ and $ H _{3}PO _{4} $ 

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. $ 1:1 $

  2. $ 1:2 $

  3. $ 2:1 $

  4. $ 2:3 $

Show answer
Correct Option: B
Explanation:

$H _{2}SO _{4}+2NaOH\rightarrow Na _{2}SO _{4}+2H _{2}O$

$H _{3}PO _{4}+NaOH\rightarrow NaH _{2}PO _{4}+H _{2}O$

Equivalent of alkali $= 19$ eq of $H _{2}SO _{4}= 1g$ eq of $H _{3}PO _{4}$

Two acids must be reacting in the ratio of their equivalent masses

Eq. wt. of $H _{2}SO _{4}=\dfrac{98}{2}=49$

Eq. wt. of $H _{3}PO _{4}= \dfrac{98}{1}=98$

$\therefore $ ratio of masses of $H _{2}SO _{4}$ & $H _{3}PO _{4}$
$49:98=1:2$

$\Rightarrow 1:2$

A gaseous alkane is exploded with oxygen. The volume of ${O} _{2}$ for complete combustion of alkane to $C{O} _{2}$ formed is in the ratio $7:4$. The molecular formula of alkane is:

introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

  1. ${C} _{2}{H} _{6}$

  2. ${C} _{3}{H} _{8}$

  3. ${C} _{4}{H} _{10}$

  4. $C{H} _{4}$

Show answer
Correct Option: A
Explanation:

The balanced reaction is given below:


${C} _{n}{H} _{2n+2} +[n+\displaystyle\frac{n+1}{2}]{O} _{2}\rightarrow nC{O} _{2} +(n+1){H} _{2}O$

Given, 

$\displaystyle\dfrac{n+\dfrac{n+1}{2}}{n}=\dfrac{7}{4}\implies n=2$

Hence, the alkane is ${C} _{2}{H} _{6}$.

Hence, the correct option is $A$

A gaseous oxide contains 30.4% of nitrogen, one molecule of which contains one nitrogen atom. The density of the oxide relative to oxygen is:

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  1. 0.94

  2. 1.44

  3. 1.50

  4. 3.0

Show answer
Correct Option: B
Explanation:

100g of oxide contain 30.4 g of Nitrogen. 


Since 1 molecule contain one nitrogen atom

14 g Nitrogen is contained by Mol. wt of oxide.

30.4 g of nitrogen present in 100 g of oxide 

$ 1 g ------------\dfrac{100}{30.4} $

$ 14 g-------------\dfrac{100 \times 14}{30.4} $

$ \dfrac{doxide}{doxygen}=\dfrac{Moxide}{Moxygen}=\dfrac{\dfrac{1400}{30.4}}{32}=1.44 $