A sample of gas with a mass of 8.40 g occupies a volume of 4.50 L at 298 K and 1.10 atm. What is the molar mass of the gas? Show all work used to find your answer. *Best answer gets medal, fan and testimony*

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A sample of gas with a mass of 8.40 g occupies a volume of 4.50 L at 298 K and 1.10 atm. What is the molar mass of the gas? Show all work used to find your answer. *Best answer gets medal, fan and testimony*

Chemistry
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Use the ideal gas law, substitute in the definition of moles, solve for M, molar mass. Ideal gas law: \(\sf PV=nRT\) moles: \(\sf n=\dfrac{m}{M}\) \(\sf \huge PV=\dfrac{m}{M}RT\rightarrow M=\dfrac{mRT}{PV}\)
May you help me plug in the variables?
I can check it for you if you give it a try

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Ok. Give me a second.
I have: 1.10*4.50 L = n(0.08206atm)(298K) And that's all I can do. Unfortunately.
that's right! you can now solve for n, then divide the mass 8.4 g by the moles.
the mass is given in the question, it's 8.4 grams
So it would be: 1.10 atm * 298 K = 24.45388 8.4 g / 24.45388 = 0.3435
not quite, first solve this equation: 1.10*4.50 L = n(0.08206atm)(298K) divide the mass by the answer (moles)
4.95 = n(0.08206*298) 4.95 = n(24.45388) 4.95/24.45388 = 0.202421... n = 0.202421
okay, looks good! now divide the mass by n
244.5398
There you go
WOOHOO! How would I properly write with sig figs?
all the numbers used have 3 sig figs, so make the answer have 3 figures.
244. Thank you so much! You're a life saver :D
that's not quite right. look at the fourth digit, you have to round up
Oh... 245, correct?
yeah. you should give this a look https://www.physics.uoguelph.ca/tutorials/sig_fig/SIG_dig.htm
Will do.
Got it. Thanks :)
no problem!
@aaronq there is problem in this question
whats the problem?
should the answer be around 41 to 42 because we are dividing 8.4 by n
you're right, the answer is 41.4974933333333333 I guess this person plugged the numbers into the calc wrong
oh thank so much
np!
thanks for double checking lol
np!

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