Q.55. 19.5 g of fluoro acetic acid (molar mass = 78 g mol) is dissolved in 500g of water at 298 K. T

Q: What is the correct answer for this exam Chemistry Q55?

The correct answer is D. Given mass of fluoro acetic acid = 19.5 g, molar mass = 78 g/mol. Mass of water = 500 g = 0.5 kg. Depression in freezing point $\Delta T_f = 1^{\circ}$C. K$_f$ for water = 1.86 K kg mol$^{-1}$. First, calculate the molality ($m$) of the solution: Moles of fluoro acetic acid = $\frac{19.5 \text{ g}}{78 \text{ g/mol}} = 0.25$ mol. Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in

Q: Q.55. 19.5 g of fluoro acetic acid (molar mass = 78 g mol) is dissolved in 500g of water at 298 K. The depression in the freezing point of water was 1°C. What is K of fluoro acetic acid ? (For water,…

The correct answer is D. Given mass of fluoro acetic acid = 19.5 g, molar mass = 78 g/mol. Mass of water = 500 g = 0.5 kg. Depression in freezing point $\Delta T_f = 1^{\circ}$C. K$_f$ for water = 1.86 K kg mol$^{-1}$. First, calculate the molality ($m$) of the solution: Moles of fluoro acetic acid = $\frac{19.5 \text{ g}}{78 \text{ g/mol}} = 0.25$ mol. Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in

Q: What are the options for this this exam Chemistry question?

A. 10^{-3} · B. 4 \times 10^{-3} · C. 3 \times 10^{-5} · D. 3 \times 10^{-3}

Question

Q.55. 19.5 g of fluoro acetic acid (molar mass = 78 g mol) is dissolved in 500g of water at 298 K. The depression in the freezing point of water was 1°C. What is K of fluoro acetic acid ? (For water, K = 1.86 K kg mol). Assume molarity and molality to have same values.

FirstInTest · Accepted Answer

Given mass of fluoro acetic acid = 19.5 g, molar mass = 78 g/mol. Mass of water = 500 g = 0.5 kg. Depression in freezing point $\Delta T_f = 1^{\circ}$C. K$_f$ for water = 1.86 K kg mol$^{-1}$.
First, calculate the molality ($m$) of the solution: Moles of fluoro acetic acid = $\frac{19.5 	ext{ g}}{78 	ext{ g/mol}} = 0.25$ mol. Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.25 	ext{ mol}}{0.5 	ext{ kg}} = 0.5$ mol/kg.
Now, use the colligative property formula: $\Delta T_f = i K_f m$. We need to find the van't Hoff factor $i$. $1 = i 	imes 1.86 	imes 0.5 \Rightarrow i = \frac{1}{1.86 	imes 0.5} = \frac{1}{0.93} \approx 1.075$.
For a weak acid like fluoro acetic acid (HA), it dissociates as $\mathrm{HA} ightleftharpoons \mathrm{H^+} + \mathrm{A^-}$. If $\alpha$ is the degree of dissociation, then $i = 1 + \alpha$. So, $\alpha = i - 1 = 1.075 - 1 = 0.075$.
The dissociation constant $K_a$ for a weak acid is given by $K_a = \frac{[H^+][A^-]}{[HA]} = \frac{C\alpha^2}{1-\alpha}$. Since molarity and molality are assumed to be same, we can use $C = m = 0.5$ mol/L.
$K_a = \frac{0.5 	imes (0.075)^2}{1 - 0.075} = \frac{0.5 	imes 0.005625}{0.925} = \frac{0.0028125}{0.925} \approx 0.003039 = 3.039 	imes 10^{-3}$.
This matches option (4) $3 	imes 10^{-3}$.

Answer

10^{-3}

Answer

4 	imes 10^{-3}

Answer

3 	imes 10^{-5}

Answer

3 	imes 10^{-3}

Related Questionsin Chemistry

Browse Engineering Questions by Subject

Practice Other Engineering Exams

Practice More Engineering Questions