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Mixture of 1 mole of each CH(4) and C(2)...

Mixture of 1 mole of each `CH_(4)` and `C_(2)H_(6)` absorb 5 mole of `Cl_(2)` to form `C Cl_(4)` and `C_(2)Cl_(6)`. Then average molecular wt. of the gaseous mixture is

A

20

B

23

C

35

D

15

Text Solution

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The correct Answer is:
B
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A mixture of 1.0 mole of Al and 3.0 mole of Cl_(2) are allowed to react as: 2Al(s)+3Cl_(2)(g) to 2AlCl_(3)(g) . Then moles of excess reagent left unreacted is:

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Knowledge Check

  • A gaseous mixture contain CH_(4) and C_(2)H_(6) in equimolecular proportion. The weight of 2.24 litres of this mixture at NTP is

    A
    (a)4.6 g
    B
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    C
    (c )1.6 g
    D
    (d)23 g

  • Assertion: C Cl_(4) and H_(2)O are immiscible . Reason : C Cl_(4) is a polar solvent.

    A
    If both assertion and reason are true and the reason is the correct explanation of the assertion
    B
    If both the assertion and reason are true but reason is not the correct explanation of the assertion
    C
    If assertion is true but reason is false
    D
    If assertion is false but reason is true.

  • Product of the following reaction is C_(6)H_(5) - N_(2)Cl + C_(6) H_(5) - NH_(2) to

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    B
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