SolutionsLecture-3
Solutions Lecture-3
Osmotic pressure example
$0.1M=\frac{n_2}{V}$
$\frac{n_2}{W_1}=0.1m$
Solution
$\Delta T=K_bm=0.05$
As $\pi=CRT= 0.1 × 22.4 = 2.29 tm$
$C=\frac{n_2}{V}=\frac{1.26/W_2}{0.02}$
$V=0.2L$
$m_2=1.26g$
$T = 300k$
$\pi=2.57×10^{-3}bar$
$\pi=CRT=\frac{1.26}{W_2×0.2}×0.083×300$
$W_2=\frac{1.26×0.083×300}{0.2×2.57×10^{-3}}$
$W_2=61.022g/mol$
$M_2=36g$
$V=1L4$
$\pi=4.98\hspace{1mm}bar$
$\pi=CRT$
$4.98=\frac{36/180}{L}RT$
$C_6H_{12}O_6$ (72+12+96)
$\pi=1.52\hspace{1mm}bar$
$1.52 = CRT$
$C=\frac{1.52}{0.083×300K}$
Abnormal molar masses
As we know,
$\Delta T_b=K_b m$
$NaCl→ Na^+_ {aq} + Cl^-_ {aq}$
$0.1m \hspace{5 mm} 0.1m \hspace{5 mm}0.1m$
$0.05m \hspace{5 mm} 0.05 \hspace{5 mm} 0.05$
$0.1(1-\alpha) \hspace{5 mm} 0.1\alpha \hspace{5 mm} 0.1\alpha$
$2A→A_2\hspace{5 mm}$
$n(1-\alpha) \hspace{2 mm} \frac{n\alpha}{2}$
Suppose we have $\quad A_mB_n$
$\quad$ $c(1-\alpha) \hspace{2 mm}$
Then conc. of $A^+$ is $c\propto m$ & conc. of $B^-$ is $c\propto n$
$c\propto m$ $\quad$ $c\propto n$
Abnormal molar mass example
Given
$M_2=2g$, $M_1=25g$
Molar mass of $C_6H_5COOH$ =84+6+32=122 g/mol
As $C_6H_5COOH→(C_6H_5COOH)_2$
$\frac{2}{122}\frac{\alpha}{2}$ $\frac{2}{122}(1-\alpha)$
$ m=4.9×0.66(1-\frac{\alpha}{2})$
$K_f=4.9KKg mol^{-1}$
As
$1-\frac{\alpha}{2}=0.500$
$\frac{\alpha}{2}=0.500$
$m=\frac{\frac{2}{122}(1-\alpha+\frac{\alpha}{2})}{25}×1000$
$m=0.66(1-\frac{\alpha}{2})$
$\alpha=1$
Van’t hoff factor
$i=\frac{C_{exp}}{C_{th}}=\frac{\pi_{exp}}{\pi_{th}}$
$i=\frac{\Delta T_{exp}}{\Delta T_{th}}$
$=\frac{exp \hspace{1mm} Colligative \hspace{1mm} Properties}{Calculated}$
$\Delta T=K_b m$
Van’t hoff factor example
Volume of acetic acid 0.6 ml
$\rho=1.06g/ml$
$V_1=1L=1 Kg$ of water ( as density is 1g/ml)
$\Delta T_f=0.0205^{\circ}C=K_f m$
$i=?$
$CH_3COOH \leftrightarrows CH_3COO^{-}+H^{+} $
$C(1-\alpha) \qquad\qquad C\alpha \qquad\qquad C\alpha$
$K_b=\frac{[H^{+}][CH_3COO^{-}]}{[CH_3COOH]}$
$K_b=\frac{(C\alpha)^2}{C(1-\alpha)}$
$K_b=\frac{C\alpha^2}{1-\alpha}$
$[C]=\frac{0.6×1.06/60}{1kg}$
$[C]=\frac{0.6×1.06}{60}m[M]$
$[C]=1.06×10^{-2}$
$W_2=10g$
$W_1=250g$
$CH_3CH_2CHClCOOH$
$\Delta T_f=K_fm=1.86×0.326$
$HA\rarr H^{+} + A $
$C(1-\alpha)\hspace{2 mm} C\alpha \hspace{2 mm} $
$=0.645^{\circ}C$
$K_a=1.4×10^{-3}$
$=\frac{[H^{+}A^{-}]}{[HA]}$
$K_f=1.86KKg/mol$
$m=C(1+\alpha)$
$=\frac{C\alpha^2}{1-\alpha}$
$=\frac{0.326\alpha^2}{1-\alpha}$
$\alpha=0.0655$
$C=\frac{moles\hspace{1 mm}of\hspace{1 mm}solute}{weight\hspace{1 mm}of\hspace{1 mm}solvent(Kg)}$
$K_a=\frac{0.326\alpha^2}{0.9346}$
$\alpha=0.0633$
$i=1+\alpha=1.0633$
$W_2=19.5g$
$W_1=500g$
$CH_2FCOOH$
$HA\rarr H^{+} + A^{-} $
$C(1-\alpha)\hspace{2 mm} C\alpha \hspace{2 mm} C\alpha $
$\hspace{2 mm}C(1+\alpha) $
$\Delta T_f=1.0^{\circ}C=K_fm=K_fC(1+\alpha)$
$=1.86×0.5(1+\alpha)$
$i=\frac{C(1+\alpha)}{C}=(1+\alpha)$
$1+\alpha=\frac{1.0}{1.86×0.5}=1.075$
$C=\frac{19.5/78}{0.5}$
$=\frac{19.5}{78×0.5}-\frac{1}{2}$
$=0.05m$
$\alpha=0.75$
$K_a=\frac{(C\alpha)^2}{C(1-\alpha)}$
$=\frac{0.5×0.075^2}{0.925}$
$=3.0×10^{-3}$
Thank you