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117 g of NaCl is added to 222 g of water in a saucepan. At what does temperature does water boil at 101.325 kPa? Ebullioscopy constant for water = 0.52 K kg mol^-1 and b.p. = 100°C(a) 98.3°C(b) 102.8°C(c) 104.7°C(d) 101.5°CThis question was posed to me in a national level competition.My query is from Colligative Properties and Determination of Molar Mass topic in chapter Solutions of Chemistry – Class 12

Answer»

The CORRECT option is (c) 104.7°C

To elaborate: Given,

Weight of solvent, W1 = 222 g

Weight of solute, w2 = 117 g

Kb = 0.53 K KG mol^-1

Now, addition of a non-volatile solute causes elevation in boiling point, ΔTb

ΔTb = (kb x 1000 x w2)/(M2 x w1)

On substituting, ΔTb = (0.52 x 1000 x 117)/(58.5 x 222) = 4.7°C

New boiling temperature = 100 + 4.7 = 104.7°C.



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