What is the ratio of the rate of diffusion of carbon(IV) oxide to that of propane, under the same conditions?

 What is the ratio of the rate of diffusion of carbon(IV) oxide to that of propane, under the same conditions?

[H= 1; C= 12; O= 16]

A. 1:1

B. 1:2

C. 2:1

D. 2:3

E. 3:1

Explanation:

The correct answer is A. 1:1

Explanation:

Let's try to understanding the concept of diffusion. 

Diffusion is the process of movement of molecules from an area of higher concentration to an area of lower concentration. 

In the context of gases, the rate of diffusion is inversely proportional to the square root of their molar masses. This is called Graham's law of diffusion.

Now, let's calculate the molar mass of carbon(IV) oxide (CO2) and propane (C3H8). The molar mass of a molecule is calculated by adding the atomic masses of all the atoms in the molecule.

The molar mass of carbon(IV) oxide (CO2) = [12 (for C) + 2*16 for two O atoms) ]= [12+32]= 44 g/mol

The molar mass of propane (C3H8) = [3*12 (for three C atoms) + 8*1 (for eight H atoms)] =[36+8]= 44 g/mol

According to Graham's law, the ratio of the rates of diffusion of two gases is given by the square root of the inverse ratio of their molar masses.

Rate of oxygen=1/√44

Rate of propane=1/√44

Rate of Oxygen/Rate of propane=1/√44÷1/√44

Rate of Oxygen/Rate of Propane=1/√44*√44/1

Rate of Oxygen/Rate of Propane=1/1

 Since the molar masses of CO2 and C3H8 are the same, the ratio of their rates of diffusion is 1:1.

So, the correct answer is Option A: 1:1. This means that carbon(IV) oxide and propane diffuse at the same rate under the same conditions.

SOURCE: NECO CHEMISTRY 2001 QUESTION 25.