1
JEE Main 2020 (Online) 5th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution is shown in the given figure. JEE Main 2020 (Online) 5th September Evening Slot Chemistry - Electrochemistry Question 65 English
The electrolyte X is :
A
HCl
B
CH3COOH
C
NaCl
D
KNO3
2
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO3 and 0.1 M AuCl. The solution was electrolyzed at 2V by passing a current of 1A for 15 minutes. The metal/metals electrodeposited will be

[ $$E_{A{g^ + }/Ag}^0$$ = 0.80 V, $$E_{A{u^ + }/Au}^0$$ = 1.69 V ]
A
Silver and gold in equal mass proportion
B
Silver and gold in proportion to their atomic weights
C
Only gold
D
Only silver
3
JEE Main 2020 (Online) 4th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
JEE Main 2020 (Online) 4th September Morning Slot Chemistry - Electrochemistry Question 68 English
$$E_{C{u^{2 + }}|Cu}^0$$ = +0.34 V

$$E_{Z{n^{2 + }}|Zn}^0$$ = -0.76 V

Identify the incorrect statement from the option below for the above cell :
A
If Eext < 1.1 V, Zn dissolves at anode and Cu deposits at cathode
B
If Eext = 1.1 V, no flow of e or current occurs
C
If Eext > 1.1 V, e flows from Cu to Zn
D
If Eext > 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode
4
JEE Main 2020 (Online) 7th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
The equation that is incorrect is :
A
$${\left( {\Lambda _m^0} \right)_{KCl}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$$
B
$${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaI}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$$
C
$${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{KCl}}$$
D
$${\left( {\Lambda _m^0} \right)_{{H_2}O}} = {\left( {\Lambda _m^0} \right)_{HCl}} + {\left( {\Lambda _m^0} \right)_{NaOH}} - {\left( {\Lambda _m^0} \right)_{NaCl}}$$
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