
Given,1 mole of substance X was treated with an excess of water, 2 moles of readily combustible gas were produced along with a solution which when reacted with\[C{{O}_{2}}\]gas produced a white turbidity. The substance X could be:
A. Ca
B. \[Ca{{H}_{2}}\]
C. \[Ca{{(OH)}_{2}}\]
D. \[Ca{{(N{{O}_{3}})}_{2}}\]
Answer
577.8k+ views
Hint: Mole is represented by mol which is the unit measuring the amount of substance in SI units; SI refers to the International System of Units. A mole of any particle or substance is defined in one another way that tells that it is exactly equal to \[6.02214076\times {{10}^{23}}\]particles and these particles may be atoms, ions, molecules or electrons.
Complete answer:
As according to question one mole of X is treated with excess of water then 2 moles of readily combustible gas were produced along with solution which when react with\[C{{O}_{2}}\]gas produced a white turbidity.
Let us discuss all the cases one by one when Ca is treated with water we get \[Ca{{(OH)}_{2}}\]only but two molecules of gases are not there so we will go for another case.
And if we take \[Ca{{(OH)}_{2}}\] and treat it with excess water then it will not give any product so this option is also wrong.
\[Ca{{(N{{O}_{3}})}_{2}}\]treated with excess of water if can form \[CaHN{{O}_{3}}\]which is not further make any white turbidity so we also mark this option wrong.
Now consider the case of \[Ca{{H}_{2}}\]treated with excess of water it gives:
$Ca{{H}_{2}}+{{H}_{2}}O\to Ca{{(OH)}_{2}}+2{{H}_{2}}$hence here 2 molecules of ${{H}_{2}}$gas is produced and further reaction gives white turbidity which is shown as:
$Ca{{(OH)}_{2}}+CO2\to CaC{{O}_{2}}+{{H}_{2}}O$; Where $CaC{{O}_{2}}$ shows white turbidity.
Hence we can say that option B is the correct answer.
Note:
Turbidity is defined as the cloudiness or haziness of a fluid caused by large numbers of individual particles that are generally invisible to the naked eye. It is similar to smoke in the air. Fluids can contain suspended solid matter consisting of particles of many different sizes. While some suspended material will be large enough and heavy enough to settle rapidly to the bottom of the container.
Complete answer:
As according to question one mole of X is treated with excess of water then 2 moles of readily combustible gas were produced along with solution which when react with\[C{{O}_{2}}\]gas produced a white turbidity.
Let us discuss all the cases one by one when Ca is treated with water we get \[Ca{{(OH)}_{2}}\]only but two molecules of gases are not there so we will go for another case.
And if we take \[Ca{{(OH)}_{2}}\] and treat it with excess water then it will not give any product so this option is also wrong.
\[Ca{{(N{{O}_{3}})}_{2}}\]treated with excess of water if can form \[CaHN{{O}_{3}}\]which is not further make any white turbidity so we also mark this option wrong.
Now consider the case of \[Ca{{H}_{2}}\]treated with excess of water it gives:
$Ca{{H}_{2}}+{{H}_{2}}O\to Ca{{(OH)}_{2}}+2{{H}_{2}}$hence here 2 molecules of ${{H}_{2}}$gas is produced and further reaction gives white turbidity which is shown as:
$Ca{{(OH)}_{2}}+CO2\to CaC{{O}_{2}}+{{H}_{2}}O$; Where $CaC{{O}_{2}}$ shows white turbidity.
Hence we can say that option B is the correct answer.
Note:
Turbidity is defined as the cloudiness or haziness of a fluid caused by large numbers of individual particles that are generally invisible to the naked eye. It is similar to smoke in the air. Fluids can contain suspended solid matter consisting of particles of many different sizes. While some suspended material will be large enough and heavy enough to settle rapidly to the bottom of the container.
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