• $0.1M=\frac{n_2}{V}$

• $\frac{n_2}{W_1}=0.1m$

Solution

• $\Delta T=K_bm=0.05$

• As $\pi=CRT= 0.1 × 22.4 = 2.29 tm$

• $0.1M=\frac{n_2}{V}$

• $C=\frac{n_2}{V}=\frac{1.26/W_2}{0.02}$

• $V=0.2L$

• $m_2=1.26g$

• $T = 300k$

• $\pi=2.57×10^{-3}bar$

• $\pi=CRT=\frac{1.26}{W_2×0.2}×0.083×300$

• $W_2=\frac{1.26×0.083×300}{0.2×2.57×10^{-3}}$

• $W_2=61.022g/mol$

• $T = 300k$

• $M_2=36g$

• $V=1L4$

• $\pi=4.98\hspace{1mm}bar$

• $\pi=CRT$

• $4.98=\frac{36/180}{L}RT$

• $C_6H_{12}O_6$ (72+12+96)

• $\pi=1.52\hspace{1mm}bar$

• $1.52 = CRT$

• $C=\frac{1.52}{0.083×300K}$

• 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 $c\propto n$

• $c\propto m$ $\quad$ $c\propto n$

• 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

• $1-\frac{\alpha}{2}=0.500$

• $\frac{\alpha}{2}=0.500$

• $m=\frac{\frac{2}{122}(1-\alpha+\frac{\alpha}{2})}{25}×1000$

• $m=0.66(1-\frac{\alpha}{2})$

• $\alpha=1$

• $i=\frac{C_{exp}}{C_{th}}=\frac{\pi_{exp}}{\pi_{th}}$

• $i=\frac{\Delta T_{exp}}{\Delta T_{th}}$

• $=\frac{exp \hspace{1mm} Colligative \hspace{1mm} Properties}{Calculated}$

• $\pi=CRT$

• $\Delta T=K_b m$

• Volume of acetic acid 0.6 ml

• $\rho=1.06g/ml$

• $V_1=1L=1 Kg$ of water ( as density is 1g/ml)

• $\Delta T_f=0.0205^{\circ}C=K_f m$

• $i=?$

• $CH_3COOH \leftrightarrows CH_3COO^{-}+H^{+}$

• $C(1-\alpha) \qquad\qquad C\alpha \qquad\qquad C\alpha$

• $K_b=\frac{[H^{+}][CH_3COO^{-}]}{[CH_3COOH]}$

• $K_b=\frac{(C\alpha)^2}{C(1-\alpha)}$

• $K_b=\frac{C\alpha^2}{1-\alpha}$

• $[C]=\frac{0.6×1.06/60}{1kg}$

• $[C]=\frac{0.6×1.06}{60}m[M]$

• $[C]=1.06×10^{-2}$

• $W_2=10g$

• $W_1=250g$

• $CH_3CH_2CHClCOOH$

• $\Delta T_f=K_fm=1.86×0.326$

• $HA\rarr H^{+} + A$

• $C(1-\alpha)\hspace{2 mm} C\alpha \hspace{2 mm}$

• $=0.645^{\circ}C$

• $K_a=1.4×10^{-3}$

• $=\frac{[H^{+}A^{-}]}{[HA]}$

• $K_f=1.86KKg/mol$

• $m=C(1+\alpha)$

• $=\frac{C\alpha^2}{1-\alpha}$

• $=\frac{0.326\alpha^2}{1-\alpha}$

• $\alpha=0.0655$

• $C=\frac{moles\hspace{1 mm}of\hspace{1 mm}solute}{weight\hspace{1 mm}of\hspace{1 mm}solvent(Kg)}$

• $K_a=\frac{0.326\alpha^2}{0.9346}$

• $\alpha=0.0633$

• $i=1+\alpha=1.0633$

• $W_2=19.5g$

• $W_1=500g$

• $CH_2FCOOH$

• $HA\rarr H^{+} + A^{-}$

• $C(1-\alpha)\hspace{2 mm} C\alpha \hspace{2 mm} C\alpha$

• $\hspace{2 mm}C(1+\alpha)$

• $\Delta T_f=1.0^{\circ}C=K_fm=K_fC(1+\alpha)$

• $=1.86×0.5(1+\alpha)$

• $i=\frac{C(1+\alpha)}{C}=(1+\alpha)$

• $1+\alpha=\frac{1.0}{1.86×0.5}=1.075$

• $C=\frac{19.5/78}{0.5}$

• $=\frac{19.5}{78×0.5}-\frac{1}{2}$

• $=0.05m$

• $\alpha=0.75$

• $K_a=\frac{(C\alpha)^2}{C(1-\alpha)}$

• $=\frac{C\alpha^2}{1-\alpha}$

• $=\frac{0.5×0.075^2}{0.925}$

• $=3.0×10^{-3}$