Question

How many grams of CO(g) are in 0.896 L of the gas at 1.48 atm and 263. K?

Answer 1

we have:

P = 1.48 atm

V = 0.896 L

T = 263.0 K

find number of moles using:

P * V = n*R*T

1.48 atm * 0.896 L = n * 0.08206 atm.L/mol.K * 263 K

n = 6.144*10^-2 mol

Molar mass of CO = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

we have below equation to be used:

mass of CO,

m = number of mol * molar mass

= 6.144*10^-2 mol * 28.01 g/mol

= 1.721 g

Answer: 1.72 g