1
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
For the following cell,

$$Zn\left( s \right)\left| {ZnS{O_4}\left( {aq} \right)} \right|\left| {CuS{O_4}\left( {aq} \right)} \right|Cu\left( s \right)$$

when the concentration of $$Z{n^{2 + }}$$ is $$10$$ times the concentration of $$C{u^{2 + }},$$ the expression for $$\Delta G$$ (in $$J\,mo{l^{ - 1}}$$) is [$$F$$ is Faraday constant; $$R$$ is gas constant; $$T$$ is temperature; $${E^0}$$ (cell)$$=1.1$$ $$V$$]
A
$$1.1F$$
B
$$2.303RT-2.2F$$
C
$$2.303RT+1.1F$$
D
$$-2.2F$$
2
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
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 12 English Option 1
B
JEE Advanced 2017 Paper 2 Offline Chemistry - Solutions Question 12 English Option 2
C
JEE Advanced 2017 Paper 2 Offline Chemistry - Solutions Question 12 English Option 3
D
JEE Advanced 2017 Paper 2 Offline Chemistry - Solutions Question 12 English Option 4
3
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
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
4
JEE Advanced 2017 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
The order of the oxidation state of the phosphorous atom in

$${H_3}P{O_2},{H_3}P{O_4},{H_3}P{O_3}$$ and $${H_4}{P_2}{O_6}$$ is
A
$${H_3}P{O_3} > {H_3}P{O_2} > {H_3}P{O_4} > {H_4}{P_2}{O_6}$$
B
$${H_3}P{O_4} > {H_3}P{O_2} > {H_3}P{O_3} > {H_4}{P_2}{O_6}$$
C
$${H_3}P{O_4} > {H_4}{P_2}{O_6} > {H_3}P{O_3} > {H_3}P{O_2}$$
D
$${H_3}P{O_2} > {H_3}P{O_3} > {H_4}{P_2}{O_6} > {H_3}P{O_4}$$
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