1.

Calculate the boiling point of 1Maqueous solution(density 1.04 g mL-) of potassiumchloride (K0.52 K kg mol)

Answer»

Molarity of solution = 1 MMolar mass of KCl(Mb)= 39 + 35.5 = 74.5 g mol-1van’t Hoff factor,i= 2 ... (AsKCldissociates completely, number of ions produced are 2.)Let volume of solution = 1 L = 1000 mLMass of KCl solution = Volume of solution × Density of solution = 1000 × 1.04 = 1040 gMolarity = No of moles of solute (nb)/ Volume of solution (V⇒nb= 1× 1 = 1Mass 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


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