પાછલા વર્ષનો NEET પ્રશ્ન-ઉકેલ L-9

પ્રશ્ન: $100^{\circ} \mathrm{C}$ પર, $6.5 \mathrm{~g}$ રસાના રસપ્રદ $100 \mathrm{~g}$ પાણીમાં ઉપલબ્ધ વેપર દબાણ $732 \mathrm{~mm}$ છે. $\mathrm{K}_{\mathrm{b}}=0.52$ હોય, તો આ રસાનો ઉક્તિ બિંદુ તરફ $102^{\circ} \mathrm{C}$ હશે#

A) $102^{\circ} \mathrm{C}$

B) $103^{\circ} \mathrm{C}$

C) $101^{\circ} \mathrm{C}$

D) $100^{\circ} \mathrm{C}$

જવાબ: $101^{\circ} \mathrm{C}$#

ઉકેલ:#

આપેલ છે કે
$$ \begin{aligned} & \mathrm{W}{\mathrm{S}}=6.5 \mathrm{~g}, \mathrm{~W}{\mathrm{A}}=100 \mathrm{~g} \ & \mathrm{p}{\mathrm{S}}=732 \mathrm{~mm} \text { of } \mathrm{Hg} \ & \mathrm{k}{\mathrm{b}}=0.52, \mathrm{~T}{\mathrm{b}}^{\mathrm{O}}=100^{\circ} \mathrm{C} \ & \mathrm{p}^0=760 \mathrm{~mm} \text { of } \mathrm{Hg} \ & \frac{p^o-p_s}{p^o}=\frac{n_2}{n_1} \ & \Rightarrow \frac{760-732}{760}=\frac{n_2}{\frac{100}{18}} \ & \Rightarrow \mathrm{n}2=0.2046 \mathrm{~mol} \ & \Delta \mathrm{T}{\mathrm{b}}=\mathrm{K}{\mathrm{b}} \times \mathrm{m} \ & \mathrm{T}{\mathrm{b}}-\mathrm{T}{\mathrm{b}}^{\circ}=k_b \times \frac{n_2 \times 1000}{w_{A(g)}} \ & \Rightarrow \mathrm{T}{\mathrm{b}}-100^{\circ} \mathrm{C}=\frac{0.52 \times 0.2046 \times 1000}{100}=1.06 \ & \Rightarrow \mathrm{T}{\mathrm{b}}=101.06^{\circ} \mathrm{C} \end{aligned} $$