Question

Match Column - X with Column – Y
$\begin{array}{cc}\text{Column-X}& \text{Column-Y}\\ \text{(p) Vapour density}& \text{(i) Unitless}\\ \text{(q) Mole}& \text{(ii) 1 mol of electrons}\\ \text{(r) 12 g Carbon}& \text{(iii) Collection of}6.023\times {10}^{23}\text{atoms}\\ \text{(s)}96500\text{C}& \text{(iv) Molecular mass (Numerical value)}\times \frac{1}{2}\end{array}$

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

Answer 1

The correct option is C (p – i, iv); (q –iii); (r –iii); (s –ii)

(a) $\text{Vapour density}=\frac{\text{Molecular mass or average molecular mass}}{massofhydrogen}$
$V.D=\frac{\text{molecular mass or average molecular mass}}{2}$
It has no units.
(b) 1 mol = $6.023\times {10}^{23}$ atoms.
(c) One mole of Carbon weighs $12\text{g}$ which means $6.023\times {10}^{23}$ atoms.
(d) $1\text{Faraday}=96500\text{C}=1\text{mol}$ of electrons.