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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).
2 years ago

Arun
23330 Points
							Dear student Given,Molarity of solution = 1 MMolar mass of KCl (Mb)= 39 + 35.5 = 74.5 g mol-1van’t Hoff factor, i = 2             ... (As KCl dissociates 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 of KCl (Wb)  = 1 × 74.5 = 74.5 g∴Mass of solvent (Wa) = 1040 – 74.5 = 965.5 g = 0.9655 kgMolality 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 RegardsArun (askIITians forum expert)

2 years ago
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