Chemistry - Mole Concept Question with Solution | TestHub

(i)$18 \mathrm{mL}$water:

As${\mathrm{d}}_{{\mathrm{H}}_2\mathrm{O}}=1\mathrm{g}/\mathrm{mL}$So${\mathrm{W}}_{{\mathrm{H}}_2\mathrm{O}}=18\mathrm{g}$

${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=\frac{18}{18}=1$

Molecules$=1\times {\mathrm{N}}_{\mathrm{A}}$

(ii)$0.18 \mathrm{g}$of water:

${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=\frac{0.18}{18}=0.01$

${\mathrm{H}}_2\mathrm{O}$$\text{molecules}=0.01\times {\mathrm{N}}_{\mathrm{A}}$

(iii)${\left({\mathrm{V}}_{{\mathrm{H}}_2\mathrm{O} \left(\mathrm{g}\right)}\right)}_{\mathrm{STP}}=0.00224\mathrm{L}$

${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}=\frac{\mathrm{V}}{22.4}=\frac{0.00224}{22.4}=0.0001$

Molecules$=0.0001\times {\mathrm{N}}_{\mathrm{A}}$

(iv)${\mathrm{n}}_{{\mathrm{H}}_2\mathrm{O}}={10}^{-3}$

${\mathrm{H}}_2\mathrm{O}$$\text{molecules}={10}^{-3}\times {\mathrm{N}}_{\mathrm{A}}$