[ આપેલ : પાણીનો મોલલ ઉત્કલનબિંદુ ઉન્નયન અચળાંક $\left(\mathrm{K}_{\mathrm{b}}\right)=0.52 \mathrm{~K} . \mathrm{kg} \mathrm{mol}^{-1}$,
$1 \mathrm{~atm}$ દબાણ $=760 \mathrm{~mm} \mathrm{Hg}$, પાણીનું મોલર દળ $\left.=18 \mathrm{~g} \mathrm{~mol}^{-1}\right]$
$\mathrm{m}=\frac{2}{0.52}$
According to question, solution is much diluted
$\text { so } \frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}=\frac{\mathrm{n}_{\text {solate }}}{\mathrm{n}_{\text {soltron: }}}$
$\frac{\Delta \mathrm{P}}{\mathrm{P}^{\circ}}=\frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text {solvent }}$
$\Delta \mathrm{P}=\mathrm{P}^{\circ} \times \frac{\mathrm{m}}{1000} \times \mathrm{M}_{\text {sohtwat }}$
$=760 \times \frac{\frac{2}{0.52}}{1000} \times 18=52.615$
$P_5=760-52.615=707.385 \mathrm{~mm} \text { of } \mathrm{Hg}$
[આપેલ $K_b (H_2O) = 0.52\, K\, kg\, mol^{-1}]$
$(Image)$
$x=$.. . . . . .(નજીક નો પૂર્ણાક)
[આપેલ : $273.15 \mathrm{~K}$ પર પાણીનો મોલલ ઠારણ બિંદુ અવનયન અયળાંક $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ છે]