Question #3979d

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Question #3979d

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Answer 1 · 2015-09-20T07:50:10

$M M = \frac{m}{P V} R T$ $MM=m/(PV)RT$

Explanation:

We can modify the ideal gas law expression $P V = n R T$ $PV = nRT$ by replacing the number of mole by $n = \frac{m}{M M}$ $n =m/(MM)$ therefore, $P V = \frac{m}{M M} R T$ $PV=m/(MM) RT$ .

Then the molar mass could be calculated from $M M = \frac{m}{P V} R T$ $MM=m/(PV)RT$ .

You plug in your numbers, you will find the answer.

$m = 0.4 g$ $m=0.4g$
$P=100xx10^3cancel(Pa)xx(1atm)/(10^5cancel(Pa)) = 1 atm$
$V=227 cm^3 = 0.227 L$
$T=27^@ C = 300K$

$MM=(0.4g)/(1cancel(atm)xx227xx10^(-3)cancel(L))xx0.08206 (cancel(L)*cancel(atm))/(mol*cancel(K))xx300cancel(K)$

$MM=43.38 g/(mol)$

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Question #3979d

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