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Calculate the boiling point of 1M aq. solution (density=1.04 gm/L) of potassium chloride (Kb =0.52 K kg /mol).

Calculate the boiling point of 1M aq. solution (density=1.04 gm/L) of potassium chloride (Kb =0.52 K kg /mol).

Grade:12

2 Answers

Arun
25750 Points
6 years ago
Dear student
 
Given,
Molarity of solution = 1 M
Molar mass of KCl (Mb)= 39 + 35.5 = 74.5 g mol-1
van’t Hoff factor, i = 2             ... (As KCl dissociates completely, number of ions produced are 2.)
Let volume of solution = 1 L = 1000 mL
Mass of KCl solution = Volume of solution × Density of solution = 1000 × 1.04 = 1040 g
Molarity = No of moles of solute (nb)/ Volume of solution (V
⇒nb = 1 × 1 = 1
Mass of KCl (Wb)/ Molar mass of KCl (Mb)  = 1
⇒ Mass of KCl (Wb)  = 1 × 74.5 = 74.5 g
∴Mass of solvent (Wa) = 1040 – 74.5 = 965.5 g = 0.9655 kg

Molality of the solution (m) =  No of moles of solute/ Mass of solvent (kg)
                                             =  1/  0.9655 = 1.03547 m
 
Delta Tb = i * Kb * m
= 2 * 0.52 * 1.0357 = 1.078 °c
 
Therefore boiling point of solution = 100 + 1.078
= 101.078°c
 
Regards
Arun (askIITians forum expert)
Yash Chourasiya
askIITians Faculty 256 Points
3 years ago
Dear Student

Given,
Molarity of solution = 1 M
Molar mass of KCl(Mb) = 39 + 35.5 = 74.5 g mol-1
van’t Hoff factor,i= 2 ... (AsKCldissociates completely, number of ions produced are 2.)
Let volume of solution = 1 L = 1000 mL
Mass of KCl solution = Volume of solution × Density of solution = 1000 × 1.04 = 1040 g
Molarity = No of moles of solute (nb)/ Volume of solution (V
⇒nb= 1× 1 = 1
Mass of KCl (Wb)/ Molar mass of KCl (Mb) = 1
⇒Mass ofKCl(Wb) = 1× 74.5 = 74.5 g
∴Mass of solvent (Wa) = 1040 – 74.5 = 965.5 g = 0.9655 kg

Molality of the solution (m) = No of moles of solute/ Mass of solvent (kg)
= 1/ 0.9655 = 1.03547 m

Delta Tb = i * Kb * m
= 2 * 0.52 * 1.0357 = 1.078 °c

Therefore boiling point of solution = 100 + 1.078
= 101.078°c

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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