
0.24 g of a volatile gas upon vaporisation gives 45 ml vapour at NTP. What will be the vapour density of the substance? (density of ${{\text{H}}_{\text{2}}}$ = 1)
A.95.39
B.5.973
C.95.93
D.59.73
Answer
582.6k+ views
Hint: Vapour density is the density of vapour relative to the density of hydrogen at constant temperature and pressure. Condition NTP indicates normal temperature (${\text{273 K}}$ ) and pressure (${\text{1 atm}}$). Use the density equation related to the molar mass of gas and molar mass of ${{\text{H}}_{\text{2}}}$gas.
Step by step answer:
We know that,
${\text{Molar mass = }}\dfrac{{{\text{mass}}}}{{{\text{mol}}}}$
\[{\text{Vapour Density = }}\dfrac{{{\text{molar mass of gas}}}}{{{\text{molar mass of }}{{\text{H}}_{\text{2}}}{\text{ gas}}}}\]
At NTP condition 1 mole of any gas has volume 22400ml.
Using this relation calculate the moles of gas present in 45ml.
${\text{45 ml of gas }} \times \dfrac{{1{\text{ mol gas}}}}{{22400{\text{ ml}}}}{\text{ = 0}}{\text{.002009 mol gas}}$
Now, calculate the molar mass of gas using the given mass of gas and calculated moles of gas.
Mass of gas = 0.24g
${\text{Molar mass = }}\dfrac{{{\text{mass}}}}{{{\text{mol}}}}$
${\text{Molar mass = }}\dfrac{{{\text{0}}{\text{.24 g}}}}{{{\text{0}}{\text{.002009 mol}}}}{\text{ = 119}}{\text{.46 g/mol}}$
So, the molar mass of the gas is ${\text{119}}{\text{.46 g/mol}}$.
Now, using the molar mass of gas and molar mass of hydrogen gas calculate the vapour density of gas as follows:
Molar mass of ${{\text{H}}_{\text{2}}}$ gas =${\text{2}}{\text{.0 g/mol}}$
\[{\text{Vapour density = }}\dfrac{{{\text{molar mass of gas}}}}{{{\text{molar mass of }}{{\text{H}}_{\text{2}}}{\text{ gas}}}}\]
\[{\text{Vapour density = }}\dfrac{{119.46{\text{ g/mo}}l}}{{2.0{\text{ g/mo}}l{\text{ }}}}\]
\[{\text{Vapour density = 59}}{\text{.73}}\]
So, the vapour density of the substance is 59.73
Thus, the correct option is (d) 59.73
Note: Vapour density is the unit less quantity. Vapour density is half of the molar mass of gas. It can also be calculated by taking the ratio of the weight of a gas to the weight of the same volume of hydrogen gas.
Weight of hydrogen gas can be calculated using the density of hydrogen gas and volume of gas. Vapour density is always measured at the constant condition of temperature and pressure.
Step by step answer:
We know that,
${\text{Molar mass = }}\dfrac{{{\text{mass}}}}{{{\text{mol}}}}$
\[{\text{Vapour Density = }}\dfrac{{{\text{molar mass of gas}}}}{{{\text{molar mass of }}{{\text{H}}_{\text{2}}}{\text{ gas}}}}\]
At NTP condition 1 mole of any gas has volume 22400ml.
Using this relation calculate the moles of gas present in 45ml.
${\text{45 ml of gas }} \times \dfrac{{1{\text{ mol gas}}}}{{22400{\text{ ml}}}}{\text{ = 0}}{\text{.002009 mol gas}}$
Now, calculate the molar mass of gas using the given mass of gas and calculated moles of gas.
Mass of gas = 0.24g
${\text{Molar mass = }}\dfrac{{{\text{mass}}}}{{{\text{mol}}}}$
${\text{Molar mass = }}\dfrac{{{\text{0}}{\text{.24 g}}}}{{{\text{0}}{\text{.002009 mol}}}}{\text{ = 119}}{\text{.46 g/mol}}$
So, the molar mass of the gas is ${\text{119}}{\text{.46 g/mol}}$.
Now, using the molar mass of gas and molar mass of hydrogen gas calculate the vapour density of gas as follows:
Molar mass of ${{\text{H}}_{\text{2}}}$ gas =${\text{2}}{\text{.0 g/mol}}$
\[{\text{Vapour density = }}\dfrac{{{\text{molar mass of gas}}}}{{{\text{molar mass of }}{{\text{H}}_{\text{2}}}{\text{ gas}}}}\]
\[{\text{Vapour density = }}\dfrac{{119.46{\text{ g/mo}}l}}{{2.0{\text{ g/mo}}l{\text{ }}}}\]
\[{\text{Vapour density = 59}}{\text{.73}}\]
So, the vapour density of the substance is 59.73
Thus, the correct option is (d) 59.73
Note: Vapour density is the unit less quantity. Vapour density is half of the molar mass of gas. It can also be calculated by taking the ratio of the weight of a gas to the weight of the same volume of hydrogen gas.
Weight of hydrogen gas can be calculated using the density of hydrogen gas and volume of gas. Vapour density is always measured at the constant condition of temperature and pressure.
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