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Question

Match Column - X with Column – Y
Column-XColumn-Y(p) Vapour density(i) Unitless(q) Mole(ii) 1 mol of electrons(r) 12 g Carbon(iii) Collection of 6.023×1023 atoms(s) 96500 C(iv) Molecular mass (Numerical value) ×12

A
(p – i, iv); (q –ii); (r –iii); (s –ii)
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B
(p – i, iii); (q –ii, iv); (r –iii); (s –i)
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C
(p – i, iv); (q –iii); (r –iii); (s –ii)
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D
(p – i); (q –ii); (r –iii, iv); (s –ii)
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Solution

The correct option is C (p – i, iv); (q –iii); (r –iii); (s –ii)
(a) Vapour density=Molecular mass or average molecular massmass of hydrogen
V.D=molecular mass or average molecular mass2
It has no units.
(b) 1 mol = 6.023×1023 atoms.
(c) One mole of Carbon weighs 12 g which means 6.023×1023 atoms.
(d) 1 Faraday=96500 C=1 mol of electrons.

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