25 mL of an aqueous solution of KCl was foun | vedclass

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

Question

(A) The reaction is: $KCl + AgNO_3 \rightarrow AgCl + KNO_3$. At the equivalence point,the millimoles of $KCl$ equal the millimoles of $AgNO_3$: $n(KC

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The reaction is: $KCl + AgNO_3 ightarrow AgCl + KNO_3$. At the equivalence point,the millimoles of $KCl$ equal the millimoles of $AgNO_3$: $n(KCl) = 20 \ mL 	imes 1 \ M = 20 \ mmol = 0.02 \ mol$. Mass of the solution $= 25 \ mL 	imes 1 \ g \ mL^{-1} = 25 \ g$. Mass of $KCl$ solute $= 0.02 \ mol 	imes 74.5 \ g \ mol^{-1} = 1.49 \ g$. Mass of solvent $= 25 \ g - 1.49 \ g = 23.51 \ g = 0.02351 \ kg$. Molality $(m) = \frac{0.02 \ mol}{0.02351 \ kg} \approx 0.8507 \ mol \ kg^{-1}$. Since $KCl$ undergoes $100 \%$ ionization,the van't Hoff factor $i = 2$. Depression in freezing point $\Delta T_f = i 	imes K_f 	imes m = 2 	imes 2.0 	imes 0.8507 = 3.4028 \ K$. The nearest integer is $3$.

Similar Questions

The solution of sugar in water contains

Distilled water boils at $373.15 \ K$ and freezes at $273.15 \ K$. $A$ solution of glucose in distilled water boils at $373.202 \ K$. What is the freezing point (in $K$) of the same solution? (For water,$K_{b}=0.52 \ K \ kg \ mol^{-1}, \ K_{f}=1.86 \ K \ kg \ mol^{-1}$)

After adding a non-volatile solute, the freezing point of water decreases to $-0.186^{\circ} C$. Calculate $\Delta T_b$ if $K_f = 1.86 \text{ K kg mol}^{-1}$ and $K_b = 0.521 \text{ K kg mol}^{-1}$. (in $\text{ K}$)

An aqueous solution of urea (molar mass $= 60 \ g \ mol^{-1}$) boils at $100.18^o C$ at atmospheric pressure. If $K_f = 1.86 \ K \ kg \ mol^{-1}$ and $K_b = 0.512 \ K \ kg \ mol^{-1}$ for water,then the freezing point of the solution is equal to ..... $^o C$.

After adding a solute,the freezing point of the solution decreases to $-0.186 \ ^{\circ}C$. Calculate $\Delta T_b$ if $K_f = 1.86 \ K \ kg \ mol^{-1}$ and $K_b = 0.521 \ K \ kg \ mol^{-1}$. (Assume the freezing point of pure solvent is $0 \ ^{\circ}C$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module