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iii. State one use of benzene.
iv. Define saponification.

(f) A chloride of an element [tex]$X$[/tex] has the formula [tex]$XCl_4$[/tex] and a vapor density of 77. Calculate the relative atomic mass of [tex]$X$[/tex]. [tex]$[Cl = 35.5]$[/tex]



Answer :

Sure, let’s address each part of the question step-by-step:

### iii. One use of benzene
Benzene is used as an intermediate in the production of various chemicals. One of its primary uses is in the production of styrene, which is a precursor for making polystyrene, a common plastic material.

### iv. Definition of saponification
Saponification is a chemical reaction between a fat or oil and a base (usually sodium hydroxide) to produce glycerol and soap. This process typically involves the hydrolysis of triglycerides (fats/oils) to form free fatty acids and glycerol, which then react with the base to produce soap.

### (f) Calculation of the relative atomic mass of X in [tex]\( XCl_4 \)[/tex] with vapor density 77
Given:
- The formula of the chloride of element [tex]\( X \)[/tex] is [tex]\( XCl_4 \)[/tex].
- Vapor density = 77.
- Relative atomic mass of Chlorine ([tex]\( Cl \)[/tex]) = 35.5.

Step-by-step solution:

1. Understanding vapor density:
Vapor density is defined as the mass of a certain volume of a substance compared to the mass of an equal volume of hydrogen, where hydrogen's vapor density is set to 1.

2. Relation between vapor density and molar mass:
The molar mass of a substance can be found using the vapor density with the following relation:
[tex]\[ \text{Molar mass} = 2 \times \text{Vapor density} \][/tex]

3. Calculate the molar mass of [tex]\( XCl_4 \)[/tex]:
[tex]\[ \text{Molar mass of } XCl_4 = 2 \times 77 = 154 \][/tex]

4. Expression for molar mass of [tex]\( XCl_4 \)[/tex]:
The molar mass of [tex]\( XCl_4 \)[/tex] is the sum of the relative atomic mass of [tex]\( X \)[/tex] and four times the relative atomic mass of Chlorine (Cl).
[tex]\[ \text{Molar mass of } XCl_4 = M_X + 4 \times 35.5 \][/tex]

5. Calculate the relative atomic mass of [tex]\( X \)[/tex] ([tex]\( M_X \)[/tex]):
[tex]\[ 154 = M_X + 4 \times 35.5 \][/tex]
[tex]\[ 154 = M_X + 142 \][/tex]
[tex]\[ M_X = 154 - 142 \][/tex]
[tex]\[ M_X = 12 \][/tex]

So, the relative atomic mass of element [tex]\( X \)[/tex] is 12.