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iii. State one use of benzene.
Benzene is an important industrial solvent used in the production of various chemicals. One significant use of benzene is in the manufacture of plastics, specifically in the production of polystyrene, which is a useful material for making plastic packaging, insulation, and other products.
iv. Define saponification.
Saponification is a chemical reaction that occurs when a fat or oil (which is a triglyceride) reacts with a strong base, typically sodium hydroxide (NaOH) or potassium hydroxide (KOH), to produce glycerol and soap. This process is commonly used in the production of soaps.
(f) A chloride of an element X has the formula [tex]\( \text{XCl}_4 \)[/tex] and a vapor density of 77. Calculate the relative atomic mass of [tex]\( \text{X} \)[/tex]. [tex]\([ \text{Cl} = 35.5 ]\)[/tex]
To find the relative atomic mass of [tex]\( \text{X} \)[/tex], follow these steps:
1. Given that the vapor density (V.D.) is 77.
2. The relationship between vapor density and molar mass is given by:
[tex]\[ \text{Vapor Density} = \frac{\text{Molar Mass}}{2} \][/tex]
Therefore, the molar mass of [tex]\( \text{XCl}_4 \)[/tex] can be calculated as:
[tex]\[ \text{Molar Mass of } \text{XCl}_4 = \text{Vapor Density} \times 2 = 77 \times 2 = 154 \][/tex]
3. The molar mass of [tex]\( \text{XCl}_4 \)[/tex] is the sum of the molar mass of the element [tex]\( \text{X} \)[/tex] and four times the molar mass of chlorine ([tex]\( \text{Cl} \)[/tex]).
4. Using the given molar mass of chlorine ([tex]\( \text{Cl} \)[/tex]) as 35.5:
[tex]\[ \text{Molar Mass of } \text{XCl}_4 = \text{Molar Mass of } \text{X} + 4 \times 35.5 \][/tex]
5. Substituting the known values:
[tex]\[ 154 = \text{Molar Mass of } \text{X} + 4 \times 35.5 \][/tex]
6. Calculate the total contribution from the four chlorine atoms:
[tex]\[ 4 \times 35.5 = 142 \][/tex]
7. Subtract this value from the molar mass of [tex]\( \text{XCl}_4 \)[/tex] to find the molar mass of the element [tex]\( \text{X} \)[/tex]:
[tex]\[ \text{Molar Mass of } \text{X} = 154 - 142 = 12 \][/tex]
Therefore, the relative atomic mass of [tex]\( \text{X} \)[/tex] is 12.
iii. State one use of benzene.
Benzene is an important industrial solvent used in the production of various chemicals. One significant use of benzene is in the manufacture of plastics, specifically in the production of polystyrene, which is a useful material for making plastic packaging, insulation, and other products.
iv. Define saponification.
Saponification is a chemical reaction that occurs when a fat or oil (which is a triglyceride) reacts with a strong base, typically sodium hydroxide (NaOH) or potassium hydroxide (KOH), to produce glycerol and soap. This process is commonly used in the production of soaps.
(f) A chloride of an element X has the formula [tex]\( \text{XCl}_4 \)[/tex] and a vapor density of 77. Calculate the relative atomic mass of [tex]\( \text{X} \)[/tex]. [tex]\([ \text{Cl} = 35.5 ]\)[/tex]
To find the relative atomic mass of [tex]\( \text{X} \)[/tex], follow these steps:
1. Given that the vapor density (V.D.) is 77.
2. The relationship between vapor density and molar mass is given by:
[tex]\[ \text{Vapor Density} = \frac{\text{Molar Mass}}{2} \][/tex]
Therefore, the molar mass of [tex]\( \text{XCl}_4 \)[/tex] can be calculated as:
[tex]\[ \text{Molar Mass of } \text{XCl}_4 = \text{Vapor Density} \times 2 = 77 \times 2 = 154 \][/tex]
3. The molar mass of [tex]\( \text{XCl}_4 \)[/tex] is the sum of the molar mass of the element [tex]\( \text{X} \)[/tex] and four times the molar mass of chlorine ([tex]\( \text{Cl} \)[/tex]).
4. Using the given molar mass of chlorine ([tex]\( \text{Cl} \)[/tex]) as 35.5:
[tex]\[ \text{Molar Mass of } \text{XCl}_4 = \text{Molar Mass of } \text{X} + 4 \times 35.5 \][/tex]
5. Substituting the known values:
[tex]\[ 154 = \text{Molar Mass of } \text{X} + 4 \times 35.5 \][/tex]
6. Calculate the total contribution from the four chlorine atoms:
[tex]\[ 4 \times 35.5 = 142 \][/tex]
7. Subtract this value from the molar mass of [tex]\( \text{XCl}_4 \)[/tex] to find the molar mass of the element [tex]\( \text{X} \)[/tex]:
[tex]\[ \text{Molar Mass of } \text{X} = 154 - 142 = 12 \][/tex]
Therefore, the relative atomic mass of [tex]\( \text{X} \)[/tex] is 12.
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