MCQ
The number of $Cl^-$ ion in $333\, g$ anhydrous $CaCl_2$ will be ................ $\mathrm{N_A}$
- ✓$6$
- B$12$
- C$3$
- D$18$
Molecule $=$ mole $\times \mathrm{N}_{\mathrm{A}}$
$=3 \times 6.02 \times 10^{23}=18.06 \times 10^{23}$
$1$ molecule $\mathrm{CaCl}_{2}$ give $=2 \mathrm{Cl}^{-}$ ion
$18.06 \times 10^{23}\left(3 \mathrm{N}_{\mathrm{A}}\right) \ldots \ldots . =2 \times 3 \times \mathrm{N}_{\mathrm{A}} \mathrm{Cl}^{-} \text {ion } $
$=6 \mathrm{N}_{\mathrm{A}} \text { ion }$
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$A ( g ) \rightleftharpoons B ( g )+\frac{1}{2} C ( g )$
the relation between dissociation constant ( $K$ ), degree of dissociation $(\alpha)$ and equilibrium pressure $(p)$ is given by?