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JEE Advanced 2019 Question Paper
April Paper – Chemistry
1. Which of the following amine will be prepared by Gabriel phthalimide reaction?
a. nButylamine
b. Triethylamine
c. neoPentylamine
d. tertButylamine
Gabriel pthalimide synthesis is used to prepare 1^{0} amines only
Answer: (a)
2. Reaction of dilute HCl with Maltose gives:
a. D â€“ glucose
b. D â€“ fructose
c. D â€“ glucose and D â€“ fructose
d. D â€“ galactose
Maltose is a dissacalaride, consisting of 2 glucose units. So on hydrolysis, it gives only Dglucose.
Answer: (a)
3. The correct order of K_{b} value of the following is
a. 1 > 2 > 3
b. 1 > 3 > 2
c. 3 > 2 > 1
d. 3 > 1 > 2
k_{b} ∝ Basic strength.
Basic strength ∝ amount of hydration ∝ Availability of Lone Pairs on Nitrogen.
Combining both effects, the order of Basic strength is, 2^{0}> 3^{0}> NH_{3}
Answer: (b)
4. Write IUPAC name of the following compound
a. 3Hydroxy2methylpentanoic acid
b. 4Methyl3hydroxypentanoic acid
c. 3Hydroxy4methylpentanoic acid
d. 2Methyl2hydroxypentanoic acid
3hydroxy 4methyl Pentanoic acid
Answer:(c)
5. Compound â€˜Xâ€™ will be
Answer: (b)
6. X will be:
Answer: (c)
7.
In a_{2}o coupling reaction, the Hydrogen is taken from the less hindered para position.
Answer: (c)
8.
Two reaction occur simultaneously
a) Ether hydrolysis using HBr
b) Electrophilic addition of HBr
Answer: (a)
9. Find the compound â€˜Xâ€™ which give following test.
Neutral FeCl_{3} ve
Fehling solution ve
Iodoform reaction +ve
Grignard reagent +ve
Neutre Fecl_{3} â†’ – ve (phenol is absent)
Fehling test â†’ – ve (no CHO)
Iodoform â†’ + ve (Î± â€“ Methyl group is present)
Grignard â†’ acidic Hatom is present
Answer: (a)
10. Identify the compound â€˜Xâ€™
a. Toluidine
b. Benzamide
c. ParaCresol
d. Oleic acid
Paracresol is acidic enough to be soluble in 10% NaOH. But Oleic acid forms salt, but is insoluble due to presence & long chain of Hydrocarbon
Answer: (c)
11. Which of the following is not correct for an ideal gas as per first low of thermodynamics
a. Adiabatic Î”U = w
b. Isothermal q = w
c. cyclic q = w
d. Isochoric Î”U = q
From 1^{st} Law,
âˆ†U = q + Ï‰
In adiabatic, q = 0
Therefore,âˆ†U = Ï‰
Answer: (a)
12. In Freundlich isotherm, \(\begin{array}{l}\frac{x}{m}\propto {{P}^{a}}.\end{array} \)
Find the value of a from the following graph
a. \(\begin{array}{l}\frac{2}{3}\end{array} \)
b. \(\begin{array}{l}\frac{1}{3}\end{array} \)
c. \(\begin{array}{l}\frac{3}{2}\end{array} \)
d. 1
Freundlich Isotherm equations:
\(\begin{array}{l}\log \frac{x}{m}=\log k+\frac{1}{n}\log P\end{array} \)
\(\begin{array}{l}y=c+mx\end{array} \)
\(\begin{array}{l}m=\frac{1}{n}=\frac{2}{3}.\end{array} \)
Answer : (a)
13. In a mixture of A and B having vapour pressure of pure A and pure B are 400 mm Hg and 600 mm Hg respectively, mole fraction of B in liquid phase is 0.5. Calculate total vapour pressure and mole fraction of A and B in vapour phase
a.500, 0.4, 0.6
b.500, 0.5, 0.5
c.450, 0.4, 0.6
d.450, 0.5, 0.5
\(\begin{array}{l}{{P}_{T}}=P\,_{A\times A}^{o}+P\,_{B\times B}^{o}\end{array} \)
= 0. 5 Ã— 400 + 0.5 Ã— 600
= 500 mm of Hg
\(\begin{array}{l}\frac{1}{{{P}_{T}}}=\frac{{{Y}_{A}}}{400}+\frac{1{{Y}_{B}}}{600}\end{array} \)
\(\begin{array}{l}âˆ´ {{Y}_{A}}=0.6\end{array} \)
\(\begin{array}{l}{{Y}_{B}}=10.6=0.4\end{array} \)
Answer : (a)
14. Arrange the following set of quantum numbers having highest energy of an electron.
(p) n = 4 l = 1 m = +1
\(\begin{array}{l}s = +\frac{1}{2}\end{array} \)
(q) n = 4 l = 2 m = 1 s
\(\begin{array}{l}=\frac{1}{2}\end{array} \)
(r) n = 3 l = 2 m = 0 s
\(\begin{array}{l}=+\frac{1}{2}\end{array} \)
(s) n = 3 l = 1 m + 1 s
\(\begin{array}{l}=\frac{1}{2}\end{array} \)
a. q > r > p > s
b. q > p > r > s
c. s > p > r > q
d. s > r > p > q
We know that n is principle quantum number and l is an azimuthal quantum number.
Also, l=nâˆ’1
The set of quantum number which have highest value of n+l will have the highest energy and vice versa.
\(\begin{array}{l}E\,\alpha \,\left( n+l \right)\end{array} \)
Answer : (c)
15. A forms ccp lattice, B occupy half of the octahedral voids and â€˜Oâ€™ occupy all the tetrahedral voids. Calculate formula â€“
a. A_{2}BO_{4}
b. ABO_{4}
c. A_{2}B_{2}O
d. A_{2}B_{2}O
CCP lattice of A â†’ no of atom of A per unit cell â†’ 4.
N oct. voids = 2 N tetra.Voids. But B occupies only half of oct. voids
∴ No of atoms of B =
\(\begin{array}{l}4\times \frac{1}{2}=2\end{array} \)
No of atoms of C â†’ 4 Ã— 2 = 8
∴ The formula â†’ A_{4} B_{2} C_{8}
Answer: (a)
16. B_{2}H_{6} reacts with O_{2} and H_{2}O respectively to form
a. B_{2}O_{3}, H_{3}BO_{3}
b. \(\begin{array}{l}{{B}_{2}}{{O}_{3}},B{{H}_{{{4}^{}}}}\end{array} \)
c. HBO_{2}, H_{3}BO_{3}
d. H_{3}BO_{3}, HBO_{2}
B_{2}H_{6} reacts with O_{2} and H_{2}O to form Diborane extensively burns in oxygen to form basic oxide and B_{2}H_{6} hydrolysis to give Boric acid
\(\begin{array}{l}{{B}_{2}}{{H}_{6}}+3{{O}_{2}}\to {{B}_{2}}{{O}_{3}}+3{{H}_{2}}O\end{array} \)
\(\begin{array}{l}{{B}_{2}}{{H}_{6}}+6{{H}_{2}}O\to 2{{H}_{3}}B{{O}_{3}}+3{{H}_{2}}\end{array} \)
Answer: (a)
17. Solution of 100 ml water contains 0.73 g of Mg(HCO_{3})_{2} and 0.81 g of Ca(HCO_{3})_{2}. Calculate the hardness in terms of ppm of CaCO_{3}.
a. 10^{2} ppm
b. 10^{4} ppm
c. 5 Ã— 10^{3} ppm
d. 10^{3} ppm
Number of moles of
\(\begin{array}{l}Mg{{\left( HC{{O}_{3}} \right)}_{2}}=\frac{0.73}{146}=0.005\end{array} \)
Number of moles of
\(\begin{array}{l}Ca{{\left( HC{{O}_{3}} \right)}_{2}}=\frac{0.81}{162}=0.005\end{array} \)
\(\begin{array}{l}PPM=\frac{Total\,no.of\,mole\times 100\times {{10}^{6}}}{100}\end{array} \)
\(\begin{array}{l}=\frac{0.005+0.005\times 100\times {{10}^{6}}}{100}\end{array} \)
\(\begin{array}{l}={{10}^{4}}PPM\end{array} \)
.
Answer: (c)
18. For \(\begin{array}{l}Z{{r}_{3}}{{(P{{O}_{4}})}_{4}}\end{array} \)
the solubility product is K_{sp} and solubility is S. Find the correct relation.
a. \(\begin{array}{l}S={{\left( \frac{{{K}_{sp}}}{6912} \right)}^{1/7}}\end{array} \)
b. \(\begin{array}{l}S={{\left( \frac{{{K}_{sp}}}{216} \right)}^{1/7}}\end{array} \)
c. \(\begin{array}{l}S={{\left( \frac{{{K}_{sp}}}{216} \right)}^{1/8}}\end{array} \)
d. \(\begin{array}{l}S={{\left( \frac{{{K}_{sp}}}{912} \right)}^{1/3}}\end{array} \)
It is A_{3}B_{4 }type of salt
Hence
\(\begin{array}{l}K{{s}_{p}}={{\left( 3s \right)}^{3}}\times {{\left( 4s \right)}^{4}}\end{array} \)
\(\begin{array}{l}=27{{s}^{3}}\times 256{{s}^{4}}\end{array} \)
\(\begin{array}{l}K{{s}_{p}}=6912{{s}^{7}}\end{array} \)
\(\begin{array}{l}S=\sqrt[7]{\frac{K{{s}_{P}}}{6912}}\end{array} \)
Answer: (a)
19. Given complexes are low spin complexes,
\(\begin{array}{l}{{[V{{(CN)}_{6}}]}^{4}}\end{array} \)
\(\begin{array}{l}{{[Cr{{(N{{H}_{3}})}_{6}}]}^{2+}}\end{array} \)
\(\begin{array}{l}{{[Ru{{(N{{H}_{3}})}_{6}}]}^{3+}}\end{array} \)
\(\begin{array}{l}{{[Fe{{(CN)}_{6}}]}^{4}}\end{array} \)
Then order of magnetic moment (Î¼) for
\(\begin{array}{l}{{V}^{2+}},F{{e}^{2+}},C{{r}^{+3}},R{{u}^{3+}}\end{array} \)
is
a.
\(\begin{array}{l}{{V}^{2+}}>C{{r}^{2+}}>F{{e}^{2+}}>R{{u}^{3+}}\end{array} \)
b. \(\begin{array}{l}F{{e}^{2+}}>{{V}^{2+}}>C{{r}^{+2}}>R{{u}^{3+}}\end{array} \)
c. \(\begin{array}{l}{{V}^{2+}}>C{{r}^{2+}}>R{{u}^{3+}}>F{{e}^{2+}}\end{array} \)
d. \(\begin{array}{l}F{{e}^{2+}}>C{{r}^{3+}}>{{V}^{2+}}>R{{u}^{3+}}\end{array} \)
For
\(\begin{array}{l}{{\left[ V{{\left( CN \right)}_{6}} \right]}^{4}}\end{array} \)
Magnetic moment
\(\begin{array}{l}=\sqrt{3\left( 3+2 \right)}=\sqrt{15}BM=3.9BM\end{array} \)
\(\begin{array}{l}{{\left[ Cl{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{2+}}\end{array} \)
Magnetic moment
\(\begin{array}{l}=\sqrt{2\left( 2+2 \right)}=\sqrt{8}=2.9BM\end{array} \)
\(\begin{array}{l}{{\left[ Ru{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{+3}}\end{array} \)
Magnetic moment
\(\begin{array}{l}=\sqrt{1\left( 1+2 \right)}=\sqrt{3}=1.732\end{array} \)
\(\begin{array}{l}{{\left[ Fe{{\left( CN \right)}_{6}} \right]}^{4}}\end{array} \)
Magnetic moment =
\(\begin{array}{l}\sqrt{no\,\,unpaired}=0\end{array} \)
Hence order
\(\begin{array}{l}{{V}^{+2}}>C{{l}^{+2}}R{{u}^{+3}}>F{{e}^{+2}}\end{array} \)
Answer: (c)
20. Given \(\begin{array}{l}E{{{}^\circ }_{{{S}_{2}}O_{8}^{2}/SO_{4}^{2}}}=2.05V\end{array} \)
\(\begin{array}{l}E_{B{{r}_{2}}/B{{r}^{}}}^{o}=1.40V\end{array} \)
\(\begin{array}{l}E_{A{{u}^{3+}}/Au}^{o}=1.10V\end{array} \)
\(\begin{array}{l}E_{{{O}_{2}}/{{H}_{2}}O}^{o}=1.20V\end{array} \)
Which of the following is the strongest oxidizing agent?
a.
\(\begin{array}{l}{{S}_{2}}O_{8}^{2}\end{array} \)
b.Br_{2}
c.Au^{+3}
d.O_{2}
More is Positive standard reduction Potential more is oxidizing nature. hence
\(\begin{array}{l}{{S}_{2}}O_{8}^{2}\end{array} \)
is answer (1)
Answer: (a)
21. Assertion: Ozone is getting depleted due to CFCs
Reason: With the deplection of ozone layer more UV radiation filters into troposphere
a. Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
b. Both Assertion and Reason are true but Reason is not correct explanation of Assertion.
c. Assertion is true but Reason is false.
d. Assertion is false but Reason is true.
Ozone gets decomposed by CFCS hence Assertion is correct
Ozone depletion cause, harmful uv rays to enter troposphere hence reason is correct but reason is not correct explanation of assertion, [2]
Answer : (b)
22. In which of the following order of hydration energy correct.
a.\(\begin{array}{l}L{{i}^{+}}>N{{a}^{+}}>{{K}^{+}}>R{{b}^{+}}>C{{s}^{+}}\end{array} \)
b.\(\begin{array}{l}L{{i}^{+}}<N{{a}^{+}}<{{K}^{+}}<R{{b}^{+}}<C{{s}^{+}}\end{array} \)
c.\(\begin{array}{l}L{{i}^{+}}>N{{a}^{+}}>{{K}^{+}}>C{{s}^{+}}>R{{b}^{+}}\end{array} \)
d.\(\begin{array}{l}L{{i}^{+}}<N{{a}^{+}}<{{K}^{+}}<C{{s}^{+}}<R{{b}^{+}}\end{array} \)
Hydration energy Î± charge, Î±_{size}^{1}
âˆ´
\(\begin{array}{l}L{{i}^{+}}>N{{a}^{+}}>{{K}^{+}}>R{{b}^{+}}>C{{s}^{+}}\end{array} \)
Answer: (a)
23. Ellingham diagram is used for
a.Reduction
b.Electrolysis
c.Zone refining
d.VanArkel
It is used for estimating, the temperature at which a particular metal oxide is reduced by another metal.
Answer: (a)
24. In isoelectronic species \(\begin{array}{l}C{{l}^{}},Ar,C{{a}^{2+}}\end{array} \)
size differ due to
a.Nuclear charge
b.Electronic â€“ electronic repulsion in valence shell
c.Magnetic quantum number
d.Principal quantum number
In isoelectronic species, number of electrons are same but size decrease with increase in atomic number or nuclear charge.
Answer: (a)
25. Three mole of Ag is heated from 300 K to 1000 K. Calculate Î”H when P = 1 atm and C_{P} = 23 + 0.01 T.
a. 62 kJ/mol
b. 45 kJ/mol
c. 38 kJ/mol
d. 54 kJ/mol
\(\begin{array}{l}\Delta H=\int\limits_{{{T}_{1}}}^{{{T}_{2}}}{n\,CpdT}\end{array} \)
\(\begin{array}{l}n\int\limits_{30}^{400}{\left( 23+0.1T \right)dT}\end{array} \)
\(\begin{array}{l}=3\left[ 23T+\frac{{{0.17}^{2}}}{2} \right]_{300}^{400}\end{array} \)
\(\begin{array}{l}=61960J/mol\Rightarrow 62k{}^\circ mol\end{array} \)
Answer: (a)
26.\(\begin{array}{l}Fe{{C}_{2}}{{O}_{4}},F{{e}_{2}}{{({{C}_{2}}{{O}_{4}})}_{3}},FeS{{O}_{4}},F{{e}_{2}}{{(S{{O}_{4}})}_{3}}\end{array} \)
one mole each, will react with how many moles of \(\begin{array}{l}KMn{{O}_{4}}\end{array} \)
a.1
b.2
c.3
d.5
\(\begin{array}{l}Fe{{C}_{2}}{{O}_{4}}+KMn{{O}_{4}} \end{array} \)
→\(\begin{array}{l} {{}}F{{e}^{3+}}+C{{o}_{2}}+M{{n}^{2+}}\end{array} \)
\(\begin{array}{l}F{{e}_{2}}{{\left( {{C}_{2}}{{O}_{4}} \right)}_{3}}+MnO_{4}^{}\end{array} \)
→ \(\begin{array}{l}{{}}F{{e}^{3+}}\end{array} \)
\(\begin{array}{l}F{{e}_{3}}{{O}_{4}}\end{array} \)
→ \(\begin{array}{l}{{}}F{{e}^{3+}}\end{array} \)
âˆ´ Total:
\(\begin{array}{l}\frac{3}{3}+\frac{6}{5}+\frac{1}{5}=2\end{array} \)
27. For 2A + B > C. Find the rate low.
[A]

[B]

Initial Rate

0.05

0.05

0.045

0.10

0.05

0.09

0.20

0.10

0.72

a. R = K[A] [B]
b. R = k[A] [B]^{2}
c. R = k[A^{2}][B]
d. R = k[A]^{2} [B]^{2}
\(\begin{array}{l}{{v}_{1}}=K{{\left[ {{A}_{1}} \right]}^{x}}{{\left[ B \right]}^{4}}\end{array} \)
\(\begin{array}{l}2={{2}^{x}}\end{array} \)
\(\begin{array}{l}x=1\end{array} \)
.
\(\begin{array}{l}\frac{0.79}{0.09}={{\left( \frac{0.2}{0.1} \right)}^{1}}{{\left[ \frac{0.1}{0.05} \right]}^{4}}\end{array} \)
\(\begin{array}{l}8={{2}^{1}}{{2}^{y}}\end{array} \)
\(\begin{array}{l}{{2}^{y}}=4\end{array} \)
\(\begin{array}{l}y=2\end{array} \)
Answer: (b)
28. Which of the following lanthanoid ions are coloured?
1.Lu^{+3}
2. Pm^{+3}
3.Sm^{+3}
4.Eu^{+3}
a.
\(\begin{array}{l}L{{u}^{+3}},P{{m}^{+3}},S{{m}^{+3}}\end{array} \)
b. \(\begin{array}{l}P{{m}^{+3}},S{{m}^{+3}},E{{u}^{+3}}\end{array} \)
c. \(\begin{array}{l}L{{u}^{+3}},P{{m}^{+3}},S{{n}^{+3}}\end{array} \)
d. \(\begin{array}{l}L{{u}^{+3}},S{{m}^{+3}},E{{u}^{+3}}\end{array} \)
\(\begin{array}{l}P{{m}^{+3}},S{{m}^{+7}},E{{u}^{+7}}\end{array} \)
contain unpaired \(\begin{array}{l}f\end{array} \)
electrons. Hence coloured.
Answer: (b)
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