1
JEE Advanced 2021 Paper 2 Online
Numerical
+2
-0
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4 solution to reach the end point. Number of moles of Fe2+ present in 250 mL solution is x $$\times$$ 10$$-$$2 (consider complete dissolution of FeCl2). The amount of iron present in the sample is y% by weight.

(Assume : KMnO4 reacts only with Fe2+ in the solution

Use : Molar mass of iron as 56 g mol$$-$$1)

The value of x is ______.
2
JEE Advanced 2021 Paper 2 Online
Numerical
+2
-0
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4 solution to reach the end point. Number of moles of Fe2+ present in 250 mL solution is x $$\times$$ 10$$-$$2 (consider complete dissolution of FeCl2). The amount of iron present in the sample is y% by weight.

(Assume : KMnO4 reacts only with Fe2+ in the solution

Use : Molar mass of iron as 56 g mol$$-$$1)

The value of y is ______.
3
JEE Advanced 2021 Paper 2 Online
+3
-1
The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below :
Correct match of the C-H bonds (shown in bold) in Column J with their BDE in Column K is

Column J
Molecule
Column K
BDE (kcal $$mo{l^{ - 1}}$$)
(P) H-CH($$C{H_3}$$)$$_2$$ (i) 132
(Q) H-CH$$_2$$Ph (ii) 110
(R) H-CH=CH$$_2$$ (iii) 95
(S) H-C $$\equiv$$ CH (iv) 88
A
P - iii, Q - iv, R - ii, S - i
B
P - i, Q - ii, R - iii, S - iv
C
P - iii, Q - ii, R - i, S - iv
D
P - ii, Q - i, R - iv, S - ii
4
JEE Advanced 2021 Paper 2 Online
+3
-1
The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by s-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below :

Cl $$-$$ Cl (g) $$\to$$ Cl$$_{(g)}^ \bullet$$ + Cl$$_{(g)}^ \bullet$$ $$\Delta$$H$$^\circ$$ = 58 kcal mol$$-$$1

H3C $$-$$ Cl (g) $$\to$$ H3C$$_{(g)}^ \bullet$$ + Cl$$_{(g)}^ \bullet$$ $$\Delta$$H$$^\circ$$ = 85 kcal mol$$-$$1

H $$-$$ Cl (g) $$\to$$ H$$_{(g)}^ \bullet$$ + Cl$$_{(g)}^ \bullet$$ $$\Delta$$H$$^\circ$$ = 103 kcal mol$$-$$1
For the following reaction

CH4(g) + Cl2(g) $$\buildrel {light} \over \longrightarrow$$ CH3Cl(g) + HCl (g)

the correct statement is
A
Initiation step is exothermic with $$\Delta$$H$$^\circ$$ = $$-$$58 kcal mol$$-$$1.
B
Propagation step involving $${}^ \bullet C{H_3}$$ formation is exothermic with $$\Delta$$H$$^\circ$$ = $$-$$2 kcal mol$$-$$1.
C
Propagation step involving CH3Cl formation is endothermic with $$\Delta$$H$$^\circ$$ = +27 kcal mol$$-$$1.
D
The reaction is exothermic with $$\Delta$$H$$^\circ$$ = $$-$$25 kcal mol$$-$$1.
EXAM MAP
Medical
NEET