JEE Main 2023 — Solutions Question with Solution
From: JEE Main 2023 (Online) 13th April Evening Shift
Question
Sea water contains and by weight of solution. The normal boiling point of the sea water is _____________ (Nearest integer)
Assume ionization for both and
Given :
Molar mass of and is 58.5 and 95 respectively.
Enter your answer
Show full solutionCorrect answer: 116.075
Correct answer
116.075
Step-by-step explanation
Amount of solvent = 100 - (29.25 + 19) = 51.75 g
Now, we can calculate the boiling point elevation using the given formula:
ΔTb = [(2 × 29.25 × 1000) / (58.5 × 51.75) + (3 × 19 × 1000) / (95 × 51.75)] × 0.52
ΔTb = 16.075
The boiling point of the sea water is the normal boiling point of water plus the change in boiling point:
Boiling point of sea water = 100 °C + 16.075 °C = 116.075 °C
Rounding to the nearest integer, the normal boiling point of the sea water is approximately 116 °C.
Now, we can calculate the boiling point elevation using the given formula:
ΔTb = [(2 × 29.25 × 1000) / (58.5 × 51.75) + (3 × 19 × 1000) / (95 × 51.75)] × 0.52
ΔTb = 16.075
The boiling point of the sea water is the normal boiling point of water plus the change in boiling point:
Boiling point of sea water = 100 °C + 16.075 °C = 116.075 °C
Rounding to the nearest integer, the normal boiling point of the sea water is approximately 116 °C.
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