Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2017-Paper-2 PART II: CHEMISTRY SECTION 1 [Maximum Marks:…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2017-Paper-2 Single Correct MCQ
Published on: August 13, 2026

PART II: CHEMISTRY

SECTION 1 [Maximum Marks: 21]

• This section contains SEVEN questions

• Each question has FOUR options

A
,
B
,
C
and
D
. ONLY ONE of these four options is correct. • For each question, marks will be awarded in one of the following categories: Full Marks +3 If only the bubble corresponding to the correct option is darkened Zero Marks 0 If none of the bubbles is darkened Negative Marks -1 In all other cases Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol -1 . The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol -1 ] Among the following, the option representing change the freezing point is

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Text Solution

Verified by Experts
The correct answer is:
C

= 3 Freezing point of ethanol + water mixture = 273- 3 = 270

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