Question

calculate the volume in dm3 occupied by 75 g of methane at STP​

Answer 1

Given :

• The mass of methane gas = 75 g

To find :

• The volume occupied by 75 g of methane gas at STP in dm³

Solution :

Given mass of methane = 75 g

Molecular weight of methane = 16 g

$\boxed {\rm{n = \frac{given \: weight}{gram \: molecular \: weight} }}$

Here ,

• n is number of moles

By substituting the values ,

$: \implies \rm \: n = \frac{75}{16} \: mol \\ \\ : \implies \rm \: n = 4.687 \: mol$

At STP ,

• Temperature = 273 K
• Pressure = 1 atm

From Ideal gas equation ,

$\boxed {\rm{PV = nRT}}$

Here ,

• P is pressure
• v is volume
• n is number of moles
• T is temperature
• R is Rydberg's constant ( = 0.0821 lit.atm.mol⁻¹K⁻¹)

We have ,

• P = 1 atm
• V = ?
• n = 4.687 mol
• R = 0.0821 lit.atm.mol⁻¹.K⁻¹
• T = 273 K

By substituting the values ,

$: \implies \rm \: (1 \: atm)(V) = (4.687 \: mol)(0.0821 \: lit.atm. {mol}^{ - 1} {k}^{ - 1} )(273 \: k) \\ \\ : \implies \rm \: V \: \cancel{atm} = (4.687 \cancel{mol})(0.0821 \: lit \: \cancel{latm. {mol}^{ - 1} {K}^{ - 1}})(273 \cancel{K}) \\ \\ : \implies \rm \: V = (4.687)(0.0821 \: lit)(273) \\ \\ : \implies \rm \: V = 150.05 \: lit$

The realtion between litres and dm³ is given by ,

$\boxed {\rm{1 \: lit = 1 \: dm {}^{3} }}$

$: \implies \rm \: V= 150.05 \: lit = 150.05 \: dm {}^{3}$

∴ The Volume occupied by the 75 g of methane gas at STP in dm³ is 150.05

Answer 2

Answer:

Given :

The mass of methane gas = 75 g

To find :

The volume occupied by 75 g of methane gas at STP in dm³

Solution :

Given mass of methane = 75 g

Molecular weight of methane = 16 g

\boxed {\rm{n = \frac{given \: weight}{gram \: molecular \: weight} }}

n=

grammolecularweight

givenweight

Here ,

n is number of moles

By substituting the values ,

\begin{gathered} : \implies \rm \: n = \frac{75}{16} \: mol \\ \\ : \implies \rm \: n = 4.687 \: mol\end{gathered}

:⟹n=

16

75

mol

:⟹n=4.687mol

At STP ,

Temperature = 273 K

Pressure = 1 atm

From Ideal gas equation ,

\boxed {\rm{PV = nRT}}

PV=nRT

Here ,

P is pressure

v is volume

n is number of moles

T is temperature

R is Rydberg's constant ( = 0.0821 lit.atm.mol⁻¹K⁻¹)

We have ,

P = 1 atm

V = ?

n = 4.687 mol

R = 0.0821 lit.atm.mol⁻¹.K⁻¹

T = 273 K

By substituting the values ,

\begin{gathered} : \implies \rm \: (1 \: atm)(V) = (4.687 \: mol)(0.0821 \: lit.atm. {mol}^{ - 1} {k}^{ - 1} )(273 \: k) \\ \\ : \implies \rm \: V \: \cancel{atm} = (4.687 \cancel{mol})(0.0821 \: lit \: \cancel{latm. {mol}^{ - 1} {K}^{ - 1}})(273 \cancel{K}) \\ \\ : \implies \rm \: V = (4.687)(0.0821 \: lit)(273) \\ \\ : \implies \rm \: V = 150.05 \: lit\end{gathered}

:⟹(1atm)(V)=(4.687mol)(0.0821lit.atm.mol

−1

k

−1

)(273k)

:⟹V

atm

=(4.687

mol

)(0.0821lit

latm.mol

−1

K

−1

)(273

K

)

:⟹V=(4.687)(0.0821lit)(273)

:⟹V=150.05lit

The realtion between litres and dm³ is given by ,

\boxed {\rm{1 \: lit = 1 \: dm {}^{3} }}

1lit=1dm

3

: \implies \rm \: V= 150.05 \: lit = 150.05 \: dm {}^{3}:⟹V=150.05lit=150.05dm

3

∴ The Volume occupied by the 75 g of methane gas at STP in dm³ is 150.05