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2. (a) State Graham's law of diffusion. [1 mark]

(b) How long would it take [tex]$30 \, cm^3$[/tex] of sulphur(IV) oxide gas to diffuse through the same conditions? (Given: [tex]$H = 1$[/tex], [tex]$Cl = 35.5$[/tex], [tex]$S = 32$[/tex], [tex]$O = 16$[/tex]) [2 marks]

(c) A gas occupied a volume of [tex]$250 \, cm^3$[/tex] at [tex]$23^{\circ} C$[/tex] and 3 atmospheres. Determine its volume under these conditions. [2 marks]

Sagot :

GinnyAnsweridn
Sure, let's break down the problem step by step.

### Part (a): Stating Graham's Law of Diffusion

Graham's Law of Diffusion

Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. In mathematical terms:

[tex]\[ \frac{\text{Rate of diffusion of Gas 1}}{\text{Rate of diffusion of Gas 2}} = \sqrt{\frac{\text{Molar Mass of Gas 2}}{\text{Molar Mass of Gas 1}}} \][/tex]

Thus, gases with lower molar masses diffuse faster than gases with higher molar masses.

### Part (b): Time for 30 cm³ of Sulphur(IV) Oxide Gas to Diffuse

To calculate the time it would take for 30 cm³ of Sulphur(IV) Oxide (SO₂) gas to diffuse, we first need to find the molar mass of SO₂.

Given the atomic masses:
- Sulfur (S) = 32
- Oxygen (O) = 16

The molar mass of SO₂ is calculated as follows:
[tex]\[ \text{Molar Mass of SO₂} = 32 + 2 \times 16 = 64 \][/tex]

Now, for Graham's Law application, since the reference gas details are not specified, we cannot proceed with finding the exact time for diffusion without that data. However, we could conclude that the time for the gas to diffuse will be dependent on the molar mass ratio as per Graham's Law.

### Part (c): Determining the Number of Moles

We are given the following data:
- Volume ([tex]\(V\)[/tex]) = 250 cm³ (or 0.250 liters)
- Temperature ([tex]\(T\)[/tex]) = 23 °C (or 23 + 273.15 = 296.15 K)
- Pressure ([tex]\(P\)[/tex]) = 3 atm

We can calculate the number of moles using the Ideal Gas Law:

[tex]\[ PV = nRT \][/tex]

Where:
- [tex]\(P\)[/tex] = 3 atm
- [tex]\(V\)[/tex] = 0.250 liters
- [tex]\(R\)[/tex] = 0.0821 L·atm/(K·mol)
- [tex]\(T\)[/tex] = 296.15 K

Rearranging to solve for [tex]\(n\)[/tex] (number of moles):

[tex]\[ n = \frac{PV}{RT} \][/tex]

Now, substituting the given values into the equation:

[tex]\[ n = \frac{(3 \, \text{atm}) (0.250 \, \text{liters})}{(0.0821 \, \text{L·atm/K·mol}) (296.15 \, \text{K})} \][/tex]

Calculating this:

[tex]\[ n = 0.03084653376471868 \, \text{moles} \][/tex]

Thus, the number of moles of the gas is approximately [tex]\(0.03085\)[/tex].

### Summary of Results

(a) Graham's Law: The rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

(b) The molar mass of SO₂ is [tex]\( 64 \)[/tex].

(c) The number of moles of the gas in a 250 cm³ volume at 23°C and 3 atm is approximately [tex]\( 0.03085 \, \text{moles} \)[/tex].
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