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Al(2)(SO(4))(3).XH(2)O has 8.1% aluminiu...

`Al_(2)(SO_(4))_(3).XH_(2)O` has `8.1%` aluminium by mass. The value of `X` is:

A

`4`

B

`10`

C

`16`

D

`18`

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The correct Answer is:
To solve the problem, we need to determine the value of \( X \) in the compound \( Al_2(SO_4)_3 \cdot XH_2O \) given that it contains \( 8.1\% \) aluminum by mass. Here’s how we can approach the solution step by step: ### Step 1: Understand the Composition The compound can be expressed as: \[ \text{Molar mass of } Al_2(SO_4)_3 \cdot XH_2O = \text{Molar mass of } Al_2(SO_4)_3 + \text{Molar mass of } XH_2O \] ### Step 2: Calculate the Molar Mass of \( Al_2(SO_4)_3 \) The molar mass of \( Al_2(SO_4)_3 \) can be calculated as follows: - Aluminum (Al): \( 2 \times 27 \, \text{g/mol} = 54 \, \text{g/mol} \) - Sulfur (S): \( 3 \times 32 \, \text{g/mol} = 96 \, \text{g/mol} \) - Oxygen (O): \( 12 \times 16 \, \text{g/mol} = 192 \, \text{g/mol} \) Adding these together: \[ \text{Molar mass of } Al_2(SO_4)_3 = 54 + 96 + 192 = 342 \, \text{g/mol} \] ### Step 3: Calculate the Molar Mass of Water The molar mass of water \( H_2O \) is: \[ \text{Molar mass of } H_2O = 2 \times 1 + 16 = 18 \, \text{g/mol} \] Thus, the molar mass of \( XH_2O \) is: \[ \text{Molar mass of } XH_2O = 18X \, \text{g/mol} \] ### Step 4: Total Molar Mass of the Compound The total molar mass of \( Al_2(SO_4)_3 \cdot XH_2O \) is: \[ \text{Total molar mass} = 342 + 18X \, \text{g/mol} \] ### Step 5: Set Up the Equation for Aluminum Percentage According to the problem, aluminum constitutes \( 8.1\% \) of the total mass. The mass of aluminum in the compound is: \[ \text{Mass of Al} = 54 \, \text{g/mol} \] Thus, we can set up the equation: \[ \frac{54}{342 + 18X} \times 100 = 8.1 \] ### Step 6: Solve for \( X \) Now, we can solve for \( X \): 1. Multiply both sides by \( (342 + 18X) \): \[ 54 \times 100 = 8.1 \times (342 + 18X) \] \[ 5400 = 8.1 \times 342 + 8.1 \times 18X \] \[ 5400 = 2770.2 + 145.8X \] 2. Rearranging gives: \[ 5400 - 2770.2 = 145.8X \] \[ 2629.8 = 145.8X \] 3. Divide by \( 145.8 \): \[ X = \frac{2629.8}{145.8} \approx 18 \] ### Final Answer Thus, the value of \( X \) is \( 18 \). ---

To solve the problem, we need to determine the value of \( X \) in the compound \( Al_2(SO_4)_3 \cdot XH_2O \) given that it contains \( 8.1\% \) aluminum by mass. Here’s how we can approach the solution step by step: ### Step 1: Understand the Composition The compound can be expressed as: \[ \text{Molar mass of } Al_2(SO_4)_3 \cdot XH_2O = \text{Molar mass of } Al_2(SO_4)_3 + \text{Molar mass of } XH_2O \] ...
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P BAHADUR-MOLE AND EQUIVALENT CONCEPT-Exercise (3A) Objective problems:
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  2. Total number of electrons present in 11.2 litre of NH(3) at STP are:

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  3. Al(2)(SO(4))(3).XH(2)O has 8.1% aluminium by mass. The value of X is:

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  8. A solution required [OH^(-)]=2M. If degree of dissociation of Mg(OH)(2...

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  9. In a compound A(x)B(y):

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  10. 20 g of an acid furnishes 0.5 "mole of" H(3)O^(+) ions in its aqueous ...

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  11. Which is not a molecular formula?

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  12. 1.0 g of pure calcium carbonate was found to require 50 mL of dilute H...

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  13. 100 mL each of 0.5 N NaOH, N//5 HCl and N//10H(2)SO(4) are mixed toget...

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  14. Vapour density of a volatile substance is 4(CH(4)=1). Its molecular we...

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  15. The equivalent weight of iron in Fe(2)O would be:

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  16. 25 mL HNO(3). If the volumes are mixed with 75 mL of 4.0 M HNO(3). If ...

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  17. To what extent must a given solution containing 40 mg AgNO(3) per mL b...

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

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

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

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