2. 20.83 g. of a gas occupies 4.167 L at 79.97 kPa at 30.0 °C. What is its molecular weight?

Question

Rodney Horn

Chemistry

14 July, 03:22

2. 20.83 g. of a gas occupies 4.167 L at 79.97 kPa at 30.0 °C. What is its molecular weight?

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Answers (1)

Know the Answer?

Avah Mclaughlin · Answer 1

f. wt. = 158 g/mol

Explanation:

Given:

mass = 20.83 g

Vol = 4.167 L

Pressure = 79.97 KPa x 0300987 Atm/KPa = 0.78924 Atm

T = 30°C + 273 = 303K

R = 0.08206 L·Atm/mol·K

PV = nRT = (mass/f. wt.) RT = > f. wt. = mass x R x T / P x V

f. wt. = 20.83g x 0.08206 L·Atm/mol·K x 303K / 0.78924 Atm x 4.167L

= 158 g / mol