Question

calculate the freezing point of 1 molar aqueous solution of kcl

Answer 1

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

Answer 2

Answer:

Explanation:

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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