
For the formation of ${\text{3}}{\text{.65g}}$ of hydrogen chloride gas what volumes of hydrogen gas and chlorine gas are required at S.T.P conditions.
A.$1.12{\text{lit}},1.12{\text{lit}}$
B. $1.12{\text{lit}},2.24{\text{lit}}$
C.${\text{3}}{\text{.65lit}},1.83{\text{lit}}$
D. ${\text{1lit}},1{\text{lit}}$
Answer
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Hint: It is a simple diatomic molecule consisting of a hydrogen atom and a chlorine atom connected with a covalent single bond. Since the chlorine atom is much more electronegative than the hydrogen atom, the covalent bond between the atoms is polar.
Formula used:
No. of moles of ${\text{HCl}}$=$\dfrac{{{\text{given}}\,{\text{mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Complete step by step answer:
No. of moles of ${\text{HCl}}$=$\dfrac{{{\text{given}}\,{\text{mass}}}}{{{\text{Molar}}\,{\text{mass}}}} = \dfrac{{{\text{3}}{\text{.65g}}}}{{{\text{36}}{\text{.5g/mol}}}} = 0.1{\text{mol}}$
$\dfrac{{\text{1}}}{{\text{2}}}{{\text{H}}_{\text{2}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}} \to {\text{HCl}}$
From the given balanced equation:
$1$mole of ${\text{HCl}}$require =$\dfrac{1}{2}$ mole of Hydrogen and $\dfrac{1}{2}$ mole of chlorine
$0.01$ mole of ${\text{HCl}}$ require=$0.05$ mole of Hydrogen and $0.05$ mole of chlorine.
According to Avogadro's law,
$1$ mole of gas occupies=$22.4{\text{litre}}$of gas at STP.
Thus $0.05$ mole of each gas occupies=$\dfrac{{22.4}}{1} \times 0.05 = 1.12$ litre at STP.
Hence the correct answer is Option A.
Additional Information: In simple words, Molar volume is the Volume occupied by one mole of any substance at a given temperature and pressure. It's generally applied only to gases, where the identity of the gas doesn't affect the volume. An ideal gas is a hypothetical gas consisting of randomly moving particles that undergo fully elastic collisions. Even though there is no such thing as an ideal gas, most of the gases tend to reach these properties when their density decreases. This happens because the intermolecular distances between gas molecules are so large that they do not interact with each other as such.
Note:
The most common example to illustrate is the molar volume of a gas at STP (Standard Temperature and Pressure), which is equal to 22.4 L for 1 mole of any ideal gas at a temperature equal to 273.15 K and a pressure equal to 1.00 atm.
Formula used:
No. of moles of ${\text{HCl}}$=$\dfrac{{{\text{given}}\,{\text{mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Complete step by step answer:
No. of moles of ${\text{HCl}}$=$\dfrac{{{\text{given}}\,{\text{mass}}}}{{{\text{Molar}}\,{\text{mass}}}} = \dfrac{{{\text{3}}{\text{.65g}}}}{{{\text{36}}{\text{.5g/mol}}}} = 0.1{\text{mol}}$
$\dfrac{{\text{1}}}{{\text{2}}}{{\text{H}}_{\text{2}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}} \to {\text{HCl}}$
From the given balanced equation:
$1$mole of ${\text{HCl}}$require =$\dfrac{1}{2}$ mole of Hydrogen and $\dfrac{1}{2}$ mole of chlorine
$0.01$ mole of ${\text{HCl}}$ require=$0.05$ mole of Hydrogen and $0.05$ mole of chlorine.
According to Avogadro's law,
$1$ mole of gas occupies=$22.4{\text{litre}}$of gas at STP.
Thus $0.05$ mole of each gas occupies=$\dfrac{{22.4}}{1} \times 0.05 = 1.12$ litre at STP.
Hence the correct answer is Option A.
Additional Information: In simple words, Molar volume is the Volume occupied by one mole of any substance at a given temperature and pressure. It's generally applied only to gases, where the identity of the gas doesn't affect the volume. An ideal gas is a hypothetical gas consisting of randomly moving particles that undergo fully elastic collisions. Even though there is no such thing as an ideal gas, most of the gases tend to reach these properties when their density decreases. This happens because the intermolecular distances between gas molecules are so large that they do not interact with each other as such.
Note:
The most common example to illustrate is the molar volume of a gas at STP (Standard Temperature and Pressure), which is equal to 22.4 L for 1 mole of any ideal gas at a temperature equal to 273.15 K and a pressure equal to 1.00 atm.
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