What is the molecular weight of a gas if a 15.0 g sample has a pressure of 836 mm Hg at 25°C in a 2.00 L flask? (R= 0.0821 L atm/ mol K)

Question

Quizplus · Accepted Answer

First, convert temperature to Kelvin: 
25°C + 273.15 = 298.15 K

Then, use the ideal gas law: 
PV = nRT

Solve for n: 
n = PV/RT

n = ((836 mmHg)(2.00 L))/(0.0821 L atm/ mol K)(298.15 K) 
n = 0.111 mol

Next, use the definition of molecular weight: 
Molecular weight = mass (g) / moles 
Molecular weight = 15.0 g / 0.111 mol 
Molecular weight = 135.1 g/mol

Convert to amu: 
1 mol / 6.0221 x 10^23 = 1 g / (6.0221 x 10^23 amu) 
135.1 g/mol x (1 mol / 6.0221 x 10^23) = 2.244 x 10^-22 amu 
Round to 3 significant figures: 
= 167 amu

What Is the Molecular Weight of a Gas If a 15.0