Answer: At STP, 0.280 L of a gas weighs 0.400 g. Calculate | StudySoup

Textbook Solutions for Chemistry

Chapter 5 Problem 43P

Question

At STP, 0.280 L of a gas weighs 0.400 g. Calculate the molar mass of the gas.

Solution

Step 1 of 2

 

The goal of the problem is to calculate the molar mass of the gas.

Given:

\(\mathrm{V}=0.280 \mathrm{~L}\)

Mass of gas \(=0.400 \mathrm{~g}\).

STP is nothing but Standard Temperature and Pressure. The standard temperature is \(273 \mathrm{Kand}\) the standard pressure is 1 atm pressure.

We know, the ideal gas equation:

\(\mathrm{PV}=\mathrm{nRT}\)

First, let's find the number of moles:

\(\begin{aligned}\mathrm{n} =\frac{P V}{R T} \\=\frac{1 a t m \times 0.280 L}{0.0821 \text { L.atm } / k / \text { mol } \times 273 K} \\=0.0124 \text { moles }\end{aligned}\)

 

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full solution

Title Chemistry 11 
Author Raymond Chang
ISBN 9780073402680

Answer: At STP, 0.280 L of a gas weighs 0.400 g. Calculate

Chapter 5 textbook questions

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