Question

calculate the volume in dm3 occupied by 32 gram methane at S.T.P.​

Answer 1

★At STP volume occupied by 1 mole of any gas=22.4l

★Now number of moles in 32 grams of methane

★ No. of moles(n)=given weight/gram molecular weight.

★n=32/16=2

★Volume occupied by 2 moles of methane=2*22.4=44.8l.

Answer 2

The volume occupied by 32 gram methane at S.T.P. is $44.8dm^3$

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm  (at STP)

V = Volume of gas = ?

n = number of moles of methane = $\frac{\text {given mass}}{\text {molar mass}}=\frac{32g}{16g/mol}=2mol$

R = gas constant =$0.0821Latm/Kmol$

T =temperature =$273K$  (at STP)

$V=\frac{nRT}{P}$

$V=\frac{2mol\times 0.0821Latm/K mol\times 273K}{1atm}=44.8L=44.8dm^3$      $(1L=1dm^3)$

Thus  the volume occupied by 32 gram methane at S.T.P. is $44.8dm^3$

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