Question

2. You have a container of N20 gas. It has a volume of 250.mL and pressure of 0.85atm at 36C. How many grams of N20 gas are present?

Answer 1

Given

Volume = 250 mL = 0.250 L (1 mL = 10^-3 L)

Pressure = 0.85 atm

Temperature = 36⁰C = (36+273) K = 309 K

We know that

PV = nRT

P is the Pressure

V is the Volume

n is no of moles

R = universal gas constant = 0.08206 L atm mol^-1 K^-1

T = 309 K

So

Lets find no of moles

n = PV/RT

= (0.85 atm * 0.250 L) / (0.08206 L atm mol^-1 K^-1 * 309 K)

= 8.38 * 10^-3 moles

So

Moles of N2O = 8.38 * 10^-3 moles

we know that

moles of N2O = Mass of N2O / molar Mass of N2O

==>

Mass of N2O = moles of N2O * molar Mass of N2O

molar Mass of N2O = 44.013 g/mole

so

Mass of N2O = 8.38 * 10^-3 moles * 44.013 g/mol

= 0.369 grams N2O is present