Question

[Help, Help, Help Use equation PV = nRT]

A 300 ml sample of gas was collected at STP weighs .90g. What is the molar mass of this gas?​

Answer 1

The molar mass should be 74.62 g/mol... I’m not really sure

Answer 2

Molar mass of the gas:

`M = 67.2 \; \text{g}\cdot \text{mol}^(-1)`.

Step-by-step explanation

How many moles of gas molecules in that 300 ml of gas sample at STP?

The gas is under Standard Temperature and Pressure, STP. Each mole of an ideal gas will occupy a volume of around 22.4 L or 22,400 ml.

By Arrhenius's Law, the volume of an ideal gas is proportional to the number of gas particles in that gas.

Assuming that the gas in question here is ideal.

`n(\text{Moles of this gas}) = \frac{V(\text{This gas under STP})}{V(\text{One mole of gas under STP})} \\\phantom{n(\text{Moles of this gas})}= \frac{300\;\text{ml}}{22,700\; \text{ml} \cdot \text{mol}^(-1)}\\\phantom{n(\text{Moles of this gas})} = (3)/(224) \; \text{mol}`.

What's the molar mass of this gas?

`M = (m)/(n) = \frac{0.90\; \text{g}}{(3)/(224) \; \text{mol}} = 67.2 \; \text{g}\cdot \text{mol}^(-1)`.