1
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
Change Language
The order of basicity among the following compounds is

JEE Advanced 2017 Paper 2 Offline Chemistry - Compounds Containing Nitrogen Question 38 English
A
$${\rm I}{\rm I} > {\rm I} > {\rm I}V > {\rm I}{\rm I}{\rm I}$$
B
$${\rm I}V > {\rm I}{\rm I} > {\rm I}{\rm I}{\rm I} > {\rm I}$$
C
$${\rm I}V > {\rm I} > {\rm I}{\rm I} > {\rm I}{\rm I}{\rm I}$$
D
$${\rm I} > {\rm I}V > {\rm I}{\rm I}{\rm I} > {\rm I}{\rm I}$$
2
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
Change Language
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 $$2$$ kg $$mo{l^{ - 1}}.$$ The figures shown below represent plots of vapor pressure $$(V.P.)$$ versus temperature $$(T).$$ [molecular weight of ethanol is $$46$$ $$g$$ $$mo{l^{ - 1}}.$$ ] Among the following, the option representing change in the freezing point is
A
JEE Advanced 2017 Paper 2 Offline Chemistry - Solutions Question 18 English Option 1
B
JEE Advanced 2017 Paper 2 Offline Chemistry - Solutions Question 18 English Option 2
C
JEE Advanced 2017 Paper 2 Offline Chemistry - Solutions Question 18 English Option 3
D
JEE Advanced 2017 Paper 2 Offline Chemistry - Solutions Question 18 English Option 4
3
JEE Advanced 2017 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
Change Language
For the following compounds, the correct statement(s) with respect to nucleophilic substitution reaction is (are)

JEE Advanced 2017 Paper 2 Offline Chemistry - Basics of Organic Chemistry Question 39 English
A
$${\rm I}$$ and $${\rm I}$$$${\rm I}$$$${\rm I}$$ follow $${S_N}1$$ mechanism
B
$${\rm I}$$ and $${\rm II}$$ follow $${S_N}2$$ mechanism
C
Compound $${\rm IV}$$ undergoes inversion of configuration
D
The order of reactivity for $${\rm I},$$ $${\rm I}{\rm I}{\rm I}$$ and $${\rm IV}$$ is : $${\rm I}V > {\rm I} > {\rm I}{\rm I}{\rm I}$$
4
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
Change Language
The standard state Gibbs free energies of formation of $$C$$(graphite) and $$C$$(diamond) at $$T=298$$ $$K$$ are

$${\Delta _f}{G^0}$$ [$$C$$(graphite)] $$ = 0kJmo{l^{ - 1}}$$

$${\Delta _f}{G^0}$$ [$$C$$(diamond)] $$ = 2.9kJmo{l^{ - 1}}$$

The standard state means that the pressure should be $$1$$ bar, and substance should be pure at a given temperature. The conversion of graphite [$$C$$(graphite)] to diamond [$$C$$(diamond)] reduces its volume by $$2 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}$$ If $$C$$(graphite) is converted to $$C$$(diamond) isothermally at $$T=298$$ $$K,$$ the pressure at which $$C$$(graphite) is in equilibrium with $$C$$(diamond), is

[Useful information : $$1$$ $$J=1$$ $$kg\,{m^2}{s^{ - 2}};1\,Pa = 1\,kg\,{m^{ - 1}}{s^{ - 2}};$$ $$1$$ bar $$ = {10^5}$$ $$Pa$$]
A
$$14501$$ bar
B
$$58001$$ $$bar$$
C
$$1450$$ bar
D
$$29001$$ bar

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