A 3.0 g sample of a gas occupies a volume of 1.00L at 100C and 740 torr pressure. The molecular weight of the (Give the same ANSWER for the next TWO problems)

gas is

94.2 g/mol

Explanation:

Ideal Gases Law can useful to solve this

P . V = n . R . T

We need to make some conversions

740 Torr . 1 atm/ 760 Torr = 0.974 atm

100°C + 273 = 373K

Let's replace the values

0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K

n will determine the number of moles

(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)

n = 0.032 moles

This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?

Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol

i'm pretty sure its b i might be wrong though!

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