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Vapour density of a volatile substance i...

Vapour density of a volatile substance is `4(CH_(4)=1)`. Its molecular weight would be:

A

`8`

B

`2`

C

`64`

D

`128`

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The correct Answer is:
To find the molecular weight of a volatile substance given its vapor density, we can use the relationship between vapor density and molecular weight. ### Step-by-Step Solution: 1. **Understand the Relationship**: The molecular weight (molar mass) of a gas can be calculated using the formula: \[ \text{Molecular Weight} = 2 \times \text{Vapor Density} \] 2. **Identify Given Information**: We are given the vapor density of the substance as 4 (relative to methane, which has a molecular weight of 16 g/mol). 3. **Calculate the Actual Vapor Density**: Since the vapor density is relative to methane, we can find the actual vapor density of the substance: \[ \text{Actual Vapor Density} = \text{Relative Vapor Density} \times \text{Vapor Density of Methane} \] \[ \text{Actual Vapor Density} = 4 \times 8 = 32 \] (Note: The vapor density of methane is 8 g/mol, as it is half of its molecular weight.) 4. **Calculate the Molecular Weight**: Now, we can use the actual vapor density to find the molecular weight: \[ \text{Molecular Weight} = 2 \times \text{Actual Vapor Density} \] \[ \text{Molecular Weight} = 2 \times 32 = 64 \text{ g/mol} \] 5. **Conclusion**: The molecular weight of the volatile substance is **64 g/mol**.

To find the molecular weight of a volatile substance given its vapor density, we can use the relationship between vapor density and molecular weight. ### Step-by-Step Solution: 1. **Understand the Relationship**: The molecular weight (molar mass) of a gas can be calculated using the formula: \[ \text{Molecular Weight} = 2 \times \text{Vapor Density} ...
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P BAHADUR-MOLE AND EQUIVALENT CONCEPT-Exercise (3A) Objective problems:
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  4. The equivalent weight of iron in Fe(2)O would be:

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  7. An oxide of metal have 20% oxygen. The eq.wt. of oxide is:

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  8. How much water is to be added to dilute 10 mL of 10 N HCl to make it d...

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  9. If 250 mL of a solution contains 24.5 g H(2)SO(4) the molarity and nor...

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  10. 0.5 mole of H(2)SO(4) is mixed with 0.2 mole of Ca(OH)(2). The maximum...

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  11. A metal oxide has 40% oxygen. The equivalent weight of the metal is:

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  12. A solution contains Na(2)CO(3) and NaHCO(3). 10 mL of the solution req...

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  13. 0.7 g "of" Na(2)CO(3).xH(2)O were dissolved in water and the volume wa...

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  14. A sample of peanut oil weighing 1.5763 g is added to 25 mL "of" 0.4210...

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  15. Eq.wt. of an acid salt NaHSO(4) is:

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  16. When a metal is burnt, its weight is increased by 24%. The equivalent ...

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  17. 0.71 g of chlorine combines with certain weight of a metal giving 1.11...

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  18. How many grams of phosphoric acid would be needed to neutralise 100 g ...

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  19. 100 mL of mixture of NaOH and Na(2)SO(4) is neutralised by 10 mL of 0....

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