
Calculate the volume in litres of \[49.8{\text{ }}gram\] of \[HCl\] gas at STP (molar mass of \[HCl\] =\[1{\text{ }} atmosphere,i.e.,760{\text{ }}mm{\text{ }}Hg\].
Answer
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Hint: At standard Temperature and Pressure (STP) the volume occupied by one mole of any gas is \[22.4{\text{ }}litres\] This volume is called the standard molar volume. It is said that \[22.4{\text{ }}litres\] of any gas at STP weighs equal to its molecular mass in grams.
Complete Step by step answer: At standard temperature (STP) is an elementary concept. In the STP system, both pressure and temperature are in their typical form. Temperature of this system is considered as \[273{\text{ }}Kelvin{\text{ }}or{\text{ }}{0^o}C\] and Standard pressure is \[1{\text{ }} atmosphere,i.e.,760{\text{ }}mm{\text{ }}Hg\]. Then experimentally, one mole of any gas occupies a volume of \[22.4{\text{ }}litres\] at \[273{\text{ }}Kelvin\] and \[1{\text{ }}atmosphere\]. The equation can be expressed
(\[1{\text{ }}mole{\text{ }}of{\text{ }}gas{\text{ }}at{\text{ }}STP = 22.4{\text{ }}litres{\text{ }}of{\text{ }}gas.)\]
Now we can calculate the volume of \[22.4{\text{ }}litres\] of the gas at S.T.P from the given weight.
Given, According to the question, the weight of \[HCl\] gas at STP = \[49.8{\text{ }}gram\]
Molar mass of \[HCl\]= (\[36.5g - mo{l^{ - 1}})\]
Now \[22.4{\text{ }}litres\] of any gas at S.T.P. weighs equal to its molecular mass in grams.
\[\;\;\therefore 36.5g\;occupies\; = 22.4{\text{ }}litres{\text{ }}of{\text{ }}volume\]
∴ \[49.8{\text{ }}gram\] of \[HCl\]occupies =\[\dfrac{{49.8 \times 22.4}}{{36.5}}\] \[ = 30.56litres.\]
So the volume in litres of \[49.8{\text{ }}gram\] of \[HCl\] gas at STP = \[ = 30.56litres.\]
Note: STP conditions are often used for engineering calculations for a gas flow or gas volume is involved. When gas volume is considered at a fixed reference temperature & pressure, its 'standard volume' is directly proportional to the molar quantity. Gas density measured at STP conditions is also a good measure of the composition of that gas, another common form of STP called standard ambient temperature and pressure or S.A.T.P. is used.
Complete Step by step answer: At standard temperature (STP) is an elementary concept. In the STP system, both pressure and temperature are in their typical form. Temperature of this system is considered as \[273{\text{ }}Kelvin{\text{ }}or{\text{ }}{0^o}C\] and Standard pressure is \[1{\text{ }} atmosphere,i.e.,760{\text{ }}mm{\text{ }}Hg\]. Then experimentally, one mole of any gas occupies a volume of \[22.4{\text{ }}litres\] at \[273{\text{ }}Kelvin\] and \[1{\text{ }}atmosphere\]. The equation can be expressed
(\[1{\text{ }}mole{\text{ }}of{\text{ }}gas{\text{ }}at{\text{ }}STP = 22.4{\text{ }}litres{\text{ }}of{\text{ }}gas.)\]
Now we can calculate the volume of \[22.4{\text{ }}litres\] of the gas at S.T.P from the given weight.
Given, According to the question, the weight of \[HCl\] gas at STP = \[49.8{\text{ }}gram\]
Molar mass of \[HCl\]= (\[36.5g - mo{l^{ - 1}})\]
Now \[22.4{\text{ }}litres\] of any gas at S.T.P. weighs equal to its molecular mass in grams.
\[\;\;\therefore 36.5g\;occupies\; = 22.4{\text{ }}litres{\text{ }}of{\text{ }}volume\]
∴ \[49.8{\text{ }}gram\] of \[HCl\]occupies =\[\dfrac{{49.8 \times 22.4}}{{36.5}}\] \[ = 30.56litres.\]
So the volume in litres of \[49.8{\text{ }}gram\] of \[HCl\] gas at STP = \[ = 30.56litres.\]
Note: STP conditions are often used for engineering calculations for a gas flow or gas volume is involved. When gas volume is considered at a fixed reference temperature & pressure, its 'standard volume' is directly proportional to the molar quantity. Gas density measured at STP conditions is also a good measure of the composition of that gas, another common form of STP called standard ambient temperature and pressure or S.A.T.P. is used.
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