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#### afchick7689

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« on: September 04, 2005, 02:08:45 PM »
The question, my work, and the answer are below, i just wanted someone to check it over-- I just wanted to make sure that no changes needed to be made to the molar mass or anything because Neon is a diatomic gas...

What is the mass in grams of neon gas that occupies a volume of 5 liters at 0°C and 1.0 atm?

PV=nRT
1.0 atm * 5 L = n * (.0821 L/atm* mol/K) * (0°C+273K)
5 = n * 22.4133
.22308 moles = n

molar mass of Ne= 20.18g

.22308mol Ne |    20.18g Ne  = 4.50179 g Ne
|     1 mol Ne

= 4.5 g Ne (W/ SIG FIGS: either 4 or 5 g Ne)

thanks so much!

#### sdekivit

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« Reply #1 on: September 04, 2005, 02:13:13 PM »
The question, my work, and the answer are below, i just wanted someone to check it over-- I just wanted to make sure that no changes needed to be made to the molar mass or anything because Neon is a diatomic gas...

What is the mass in grams of neon gas that occupies a volume of 5 liters at 0°C and 1.0 atm?

PV=nRT
1.0 atm * 5 L = n * (.0821 L/atm* mol/K) * (0°C+273K)
5 = n * 22.4133
.22308 moles = n

molar mass of Ne= 20.18g

.22308mol Ne |    20.18g Ne  = 4.50179 g Ne
|     1 mol Ne

= 4.5 g Ne (W/ SIG FIGS: either 4 or 5 g Ne)

thanks so much!

the method looks fine to me

#### Yggdrasil

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