16. A 4.37 g sample of a certain diatomic gas occupies a volume of 3.00-L at 1.00 atm and a temperature of 45°C. Identify the gas. Anyone please help?

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16. A 4.37 g sample of a certain diatomic gas occupies a volume of 3.00-L at 1.00 atm and a temperature of 45°C. Identify the gas. Anyone please help?

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Answer 1 · 2015-06-03T07:04:14

Working equation: $P V = n R T$ $PV = nRT$ (Ideal gas law )
n (no of moles) = m(mass)/M (Molecular wt. )

R in Lt-atm scale 0.08205736(14)L-atm /K-mol ~ 0.082

So, fitting values
$3 \cdot 1$ $3*1$ = $(\frac{4.37}{M}) \cdot 0.082 \cdot 318$ $(4.37/M)*0.082*318$ (Temperature in Kelvin scale)
$\Rightarrow 3 = \frac{4.37}{M} \cdot 0.082 \cdot 318$ $=>3= 4.37/M*0.082*318$
$\Rightarrow 3 = \frac{113.95}{M}$ $=>3= 113.95/M$
$\Rightarrow M ~ 37.985$ $=>M ~ 37.985$

Now the only diatomic gas which has closest molecular mass is Fluorine ( $F_{2}$ $F_2$ ) (37.997). The difference in value is probably coming from approximation.

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16. A 4.37 g sample of a certain diatomic gas occupies a volume of 3.00-L at 1.00 atm and a temperature of 45°C. Identify the gas. Anyone please help?

1 Answer

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