
In a given sample of bleaching powder, the percentage of available chlorine is 35. The volume of chlorine obtained if 10 g of the sample is treated with $\text{HCl}$at NTP:
A) $1.51$
B) $3.51$
C) $151$
D) $1501$
Answer
546.6k+ views
Hint:
This question can be solved from the knowledge of the ideal gas law equation. We shall substitute the appropriate values in the equation given below to find the mass of chlorine and thus its moles. Then using the ideal gas equation, we shall calculate the volume of chlorine.
Formula Used:
The available chlorine that is liberated for the reaction of a certain amount of bleaching powder with dilute hydrochloric acid is mathematically represented as: $\dfrac{\text{mass of the chlorine produced}}{\text{weight of the sample}}\times 100$
Complete step by step solution:
When bleaching powder reacts with any kind acid, such as dilute hydrochloric acid, acetic acid, etc. then chlorine gas is liberated along with the formation of a calcium salt. The amount of chlorine gas thus liberated is called available chlorine.
The formula for the percentage of available chlorine is = $\dfrac{\text{mass of the chlorine produced}}{\text{weight of the sample}}\times 100$
According to the question, the percentage of available chlorine is 35 and the weight of the sample is 10 grams. Putting the values in the above equation we get,
$\text{mass of the chlorine produced}=\dfrac{35\times 10}{100}=3.5$ Grams.
Now, $71$grams of Chlorine gas is equal to 1 mole, therefore,
$3.5$Grams of chlorine is equal to $\dfrac{1}{71}\times 3.5=0.05$ mole of chlorine.
At N.T.P, 1 mole of chlorine \[22.4\]L, which is the molar volume occupied by every gas at N.T.P,
Therefore $0.05$moles = $22.4\times 0.05=1.12\approx 1.5$L of chlorine.
The correct answer is option A.
Note:
The molar volume of the gas is defined by the volume occupied by one mole of any gas at standard temperature and pressure and is equal to one mole of molecules or one mole of atoms of the gas and is also equal to the molecular or atomic weight of the gas.
The molar volume is applicable to any gas at standard temperature and pressure only.
This question can be solved from the knowledge of the ideal gas law equation. We shall substitute the appropriate values in the equation given below to find the mass of chlorine and thus its moles. Then using the ideal gas equation, we shall calculate the volume of chlorine.
Formula Used:
The available chlorine that is liberated for the reaction of a certain amount of bleaching powder with dilute hydrochloric acid is mathematically represented as: $\dfrac{\text{mass of the chlorine produced}}{\text{weight of the sample}}\times 100$
Complete step by step solution:
When bleaching powder reacts with any kind acid, such as dilute hydrochloric acid, acetic acid, etc. then chlorine gas is liberated along with the formation of a calcium salt. The amount of chlorine gas thus liberated is called available chlorine.
The formula for the percentage of available chlorine is = $\dfrac{\text{mass of the chlorine produced}}{\text{weight of the sample}}\times 100$
According to the question, the percentage of available chlorine is 35 and the weight of the sample is 10 grams. Putting the values in the above equation we get,
$\text{mass of the chlorine produced}=\dfrac{35\times 10}{100}=3.5$ Grams.
Now, $71$grams of Chlorine gas is equal to 1 mole, therefore,
$3.5$Grams of chlorine is equal to $\dfrac{1}{71}\times 3.5=0.05$ mole of chlorine.
At N.T.P, 1 mole of chlorine \[22.4\]L, which is the molar volume occupied by every gas at N.T.P,
Therefore $0.05$moles = $22.4\times 0.05=1.12\approx 1.5$L of chlorine.
The correct answer is option A.
Note:
The molar volume of the gas is defined by the volume occupied by one mole of any gas at standard temperature and pressure and is equal to one mole of molecules or one mole of atoms of the gas and is also equal to the molecular or atomic weight of the gas.
The molar volume is applicable to any gas at standard temperature and pressure only.
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