19.5 text{ g} of fluoroacetic acid (molar mas | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $1$. Calculate the number of moles of fluoroacetic acid: $	ext{Moles} = \frac{19.5 	ext{ g}}{78 	ext{ g mol}^{-1}} = 0.25 	ext{ mol}$.$2$. Cal

Vedclass · Accepted Answer

$3 	imes 10^{-3}$

Vedclass · Answer

(D) $1$. Calculate the number of moles of fluoroacetic acid: $	ext{Moles} = \frac{19.5 	ext{ g}}{78 	ext{ g mol}^{-1}} = 0.25 	ext{ mol}$.$2$. Calculate the molality $(m)$: $m = \frac{0.25 	ext{ mol}}{0.5 	ext{ kg}} = 0.5 	ext{ m}$.$3$. Use the freezing point depression formula: $\Delta T_f = i \cdot K_f \cdot m$.$4$. Given $\Delta T_f = 1 	ext{ K}$, $K_f = 1.86 	ext{ K kg mol}^{-1}$, and $m = 0.5 	ext{ m}$, we have: $1 = i 	imes 1.86 	imes 0.5$.$5$. Solving for the van't Hoff factor $(i)$: $i = \frac{1}{0.93} \approx 1.075$.$6$. Since $i = 1 + \alpha$ for a weak acid, $\alpha = i - 1 = 1.075 - 1 = 0.075$.$7$. Calculate the dissociation constant $(K_a)$: $K_a = C \alpha^2 = 0.5 	imes (0.075)^2 = 0.5 	imes 0.005625 = 0.0028125 \approx 3 	imes 10^{-3}$.

Similar Questions

The freezing point of a $0.05 \ m \ BaCl_2$ solution in water ($100 \%$ ionisation) is about $(K_f = 1.86 \ K \ kg \ mol^{-1})$. (in $^oC$)

If $1$ mole of a solute dissociates to give $n$ number of ions,then the degree of dissociation $\alpha$ in terms of the van't Hoff factor $i$ is given by:

What will be the osmotic pressure in $atm$ of a $0.2 \ M$ $K_4[Fe(CN)_6]$ solution at $27 \ ^oC$?

Which of the following will show maximum osmotic pressure for its $1 \, m$ solution?

When $HgI_2$ is added to an aqueous solution of $KI$,then the:

Vedclass Test Series

Exam Paper Generator

Online Exam Module