1
JEE Main 2022 (Online) 27th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Match List - I with Match List - II.

List - I List - II
(A) Glucose + HI (I) Gluconic acid
(B) Glucose + Br$$_2$$ water (II) Glucose pentacetate
(C) Glucose + acetic anhydride (III) Saccharic acid
(D) Glucose + HNO$$_3$$ (IV) Hexane

Choose the correct answer from the options given below:

A
(A) - (IV), (B) - (I), (C) -(II), (D) - (III)
B
(A) - (IV), (B) - (III), (C) -(II), (D) - (I)
C
(A) - (III), (B) - (I), (C) -(IV), (D) - (II)
D
(A) - (I), (B) - (III), (C) -(IV), (D) - (II)
2
JEE Main 2022 (Online) 27th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Match List - I with List - II.

List - I
(Mixture)
List - II
(Purification Process)
(A) Chloroform & Aniline (I) Steam distillation
(B) Benzoic acid & Napthalene (II) Sublimation
(C) Water & Aniline (III) Distillation
(D) Napthalene & Sodium chloride (IV) Crystallisation

Choose the correct answer from the options given below :

A
(A) - (IV), (B) - (III), (C) - (I), (D) - (II)
B
(A) - (III), (B) - (I), (C) - (IV), (D) - (II)
C
(A) - (III), (B) - (IV), (C) - (II), (D) - (I)
D
(A) - (III), (B) - (IV), (C) - (I), (D) - (II)
3
JEE Main 2022 (Online) 27th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

$$\mathrm{Fe}^{3+}$$ cation gives a prussian blue precipitate on addition of potassium ferrocyanide solution due to the formation of :

A
$$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]_{2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$$
B
$$\mathrm{Fe}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{2}$$
C
$$\mathrm{Fe}_{3}\left[\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{CN})_{4}\right]_{2}$$
D
$$\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$$
4
JEE Main 2022 (Online) 27th July Evening Shift
Numerical
+4
-1
Change Language

The normality of $$\mathrm{H}_{2} \mathrm{SO}_{4}$$ in the solution obtained on mixing $$100 \mathrm{~mL}$$ of $$0.1 \,\mathrm{M} \,\mathrm{H}_{2} \mathrm{SO}_{4}$$ with $$50 \mathrm{~mL}$$ of $$0.1 \,\mathrm{M}\, \mathrm{NaOH}$$ is _______________ $$\times 10^{-1} \mathrm{~N}$$. (Nearest Integer)

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