JEE Advanced 2019 Chemistry Paper 1 solutions are available here. All questions are solved by subject experts and in a detailed manner. Students can also download the solved paper in a PDF format for offline practice to improve their speed and accuracy. Practising JEE Advanced 2019 Chemistry question Paper 1 will help the aspirants to understand the overall exam pattern and prepare accordingly for upcoming exam.
Paper 1 - Chemistry
1. Which of the following set represent correct formula for Malachite, Magnetite, Calamine & Cryolite?
- a) CuCO3, Fe2O3, ZnO, Al2O3
- b) CuCO3,Cu(OH), Fe3O4, ZnCO3, Na3 AlF6
- c) CuCO3, Fe3O4, ZnCO3, Al2O3
- d) CuCO3.Cu(OH), Fe2O3, ZnCO3, Na3AlF6
Solution:
Malachite → CuCO3.Cu(OH)2
Magnetite → Fe3O4
Calamine → ZnCO3
Cryollite → Na3AlF6
Answer: (b)
2.The correct order of acid strength of the following carboxylic acids is
Solution:
- Answer is (1)
3. Sodium stearate is a strong electrolyte. Which of the following plot is correct regarding its conductance:
Solution:
By definition, λmα (1/√C)
Answer: (b)
4. Which green coloured compound of chromium is formed in borax bead test?
- a) Cr(BO2)3
- b) Cr2O3
- c) CrB
- d) CrBO3
Solution:
Answer: (a)
5. Choose the reaction, for which the standard enthalpy of reaction is equal to the standard enthalpy of formation:
Solution:
By definition,
Enthalpy of formation is defined as the Enthalpy change occurring when, a compound is formed from its constituent elements in standard state.
Answer: (B, C)
6. A Tin-chloride ‘P’ gives following reaction (unbalanced reaction)
P + Cl-
P + Me3N
P + CuCl2
Then which of the following is/are correct.
- a) Y contains co-ordinate bond
- b) X is sp3 hybridised.
- c) Oxidation state of Sn is X is +1.
- d) X contain lone pair on central atom.
Solution:
Answer: (a, b, d)
7.
x1, x2, x3, x4 are either particles or radiation. Then
- a) x1 is deflected toward negatively charged plate.
- b) x2 is β-particle.
- c) x3 is γ-radiation.
- d) z is isotope of 238U
Solution:
-
x1 -> α - decay
x2 -> β - decay
x3 -> β - decay
x4 -> α – decay
Answer: (a, b, d)
8. Fusion of MnO2 along with KOH and O2 forms X. Electrolytic oxidation of X yields Y. X undergoes disproportionation reaction in acidic medium to MnO2 and Y. The Manganese in X and Y is in the form W & Z respectively, then
- a) W & Z are coloured
- b) W is diamagnetic and Z is paramagnetic
- c) Both W & Z are tetrahedral in shape
- d) Both W & Z involve pπ-dπ bonding for π bond
Solution:
Answer: (a, c, d)
9. Choose the corection option for the following reaction
Solution:
-
Answer: (1, 4)
10. Which of the following are true.
- a) Monosachharides can not be hydrolysed to give polyhydroxy aldehydes and ketones.
- b) Hydrolysis of sucrose gives dextrorotatory glucose and laevorotatory fructose
- c) Oxidation of glucose with bromine water gives glutamic acid.
- d) The two six membered hemiacetal form of D(+) glucose are anomers.
Solution:
Answer is: (a, b, d)
11. Identify the option where all four molecules possess permanent dipole moment at room temperature.
- a) BF, O3, SF6, XeF6
- b) BeCl2, CO2, BCl3, CHCl3
- c) SO2, C6H5Cl, H2Se, BrF5
- d) NO2, NH3, POCl3, CH3Cl
Solution:
For Option c
Answer: (c, d)
12. Which of the following is/are correct regarding root mean square speed (Urms) & average translation K.E. (Eav) of molecule in a gas at equilibrium.
- a) Eav is doubled when its temperature is increased 4 times
- b) Urms is inversely proportional to the square root of its molecular mass
- c) Eav at a given temperature doesn’t depend on its molecular mass
- d) Urms is doubled when its temperature is increased 4 times
Solution:
Eav = (3/2)RT [independent of mass)
urms =
\(\sqrt{\frac{3RT}{M}}\)Answer: (b, c, d)
13. XeF4 + O2F2
Solution:
- 19
Distorted octahedral shape.
14. For the following reaction, equilibrium constant Kc at 298 K is 1.6 × 1017.
When equal volume of 0.06 M Fe+2 and 0.2 M S-2 solution are mixed, then equilibrium concentration of Fe+2 is found to be Y × 10-17 M. Y is:
Solution:
- \(Fe^{2+} + S^{2-} \rightarrow FeS \)
Ini
0.06
0.2
0
After mix
0.03
0.1
At 0 q/m
x
0.07
15.Schemes 1 and 2 describe the conversion of P to Q and R to S, respectively. Scheme 3 describes the synthesis of T from Q and S. The total number of Br atoms in a molecule of T is _____
Solution:
-
Total Br atoms = 4
16. Which of the following compounds contain bond between same type of atoms.
N2O4, B3 N3H6, H2S2O3, N2O, H2S2O8, B2H6
Solution:
Answer: N2O4, H2S2O3, N2O, H2S2O8
17. Consider the kinetic data given in the following table for the reaction A + B + C → Product
When [A] = 0.15
[B] = 0.25
[C] = 0.15
Rate of reaction is Y × 10-5 m/s. Find Y.
Solution:
-
Let
\(r = K[A]^x[B]^y[C]^z\)From (1), (2) -> y = 0
(1), (3) -> z = 1
(1), (4) -> x = 1
Therefore, rate law becomes
\(r = K[A]^1[B]^0[C]^1\)From (2)
K = 3 x 10-3
When [A] = 0.15, [B] = 0.25, [C] = 0.15
r = 3 x 10-3 [0.15]1 [0.15]1
= 6.75 x 10-5 moll-1 s-1
So, y x 10-6
Hence, y = 6.75
Answer: (6.75)
18. On dissolving 0.5 g of non-volatile, non-ionic solute to 39 g of benzene, its vapour pressure decreases from 650 mm of Hg to 640 mm of Hg. The depression of freezing point of benzene (in K) upon addition of the solute is __________________.
[Given data: Molar mass & molar freezing point depression of benzene is 78 g mol-1 & 5.12 K Kg mol-1]
Solution:
-
\(\frac{P^0 - P_S}{P_S} = \frac{n_{solute}}{n_{solvant}}\)\(\frac{650 - 640}{640} - \frac{1 \times 0.58 \times 78}{M \times 39}\)
Therefore, M = 64 g
Video Lessons - Paper 1 Chemistry
JEE Advanced 2019 Chemistry Paper 1 Solutions
