Question #3973b

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Question #3973b

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Answer 1 · 2014-11-10T05:19:53

The molar mass for the gas in the example problem is $42 \frac{g}{m o l}$ $42 g/(mol)$ .

Starting with the definition of molar mass,

$m o l a r m a s s = \frac{g r a m s}{m o l}$ $molar mass = (grams)/(mol)$

and solve for the number of moles to get the equation

$m o l = \frac{g r a m s}{m o l a r m a s s}$ $mol = (grams)/(molar mass)$ .

This equation can be substituted into the ideal gas law, $P V = n R T$ $PV = nRT$ , to get

$P V = \frac{g R T}{m o l a r m a s s}$ $PV=(gRT)/(molar mass)$

and rearranging this to solve for molar mass gives

$m o l a r m a s s = \frac{g R T}{P V}$ $molar mass=(gRT)/(PV)$

with this and some simple unit conversions, we can now calculate.

$m o l a r m a s s = \frac{1.62 g \times 0.0821 \times 293 K}{0.9842 a t m \times 0.941 L} = 42 \frac{g}{m o l}$ $molar mass = (1.62g xx 0.0821 xx 293K)/(0.9842 atm xx 0.941 L) = 42 g/(mol)$

Hope this helps.

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Question #3973b

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