Dear Student, The liberation of each H+ ion can result in one NaOH being formedNow, Q=It=nFor, n=ItF=9 65×100096500=0 1Moles of H+ liberated=0 1Therefore, Moles of NaOH formed=0 1Mass of NaOH formed=0 1×40=4 gHence, correct answer is (B)

Q.no.40

Q.no.40 A current ot 2 A was passed for 1 h through a solution of CuS04 0.237 g of Cu2* ions were discharged at cathode. The current efficiency is (A)42.2 % (B) 26.1 % Se ion (G) : Applications of Electrolysis (D) 40.01 % CO. current of 9.65 ampere is passed through the aqueous solution NaCl using suitable electrodes for 1000 s. The amount of NaOH formed during electrolysis is (A) 2.0 g (B) 4.0 g (C)6.o g (D) 8.0 g . In an electrolytic cell of Ag/AgN03/Ag, when current is passed, the concentration of AgN03 (A)increases (B) decreases )Remains same (D)None of these 42. If 0.224 L of H2 gas is formed at the cathode, the volume of 02 gas•formed at the anode (A) 0.224 L under identical conditions, is (C) 0.122 L (B) 0.448 L (D) 1.12 L Section (H) : primary Cell and Secondary Cell lead storage cell the density of sulphuric acid in the cell

Dear Student,

The liberation of each H^{+} ion can result in one NaOH being formed Now, Q = It = nF or, n = \frac{It}{F} = \frac{9.65 \times 1000}{96500} = 0.1 Moles of H^{+} liberated = 0.1 Therefore, Moles of NaOH formed = 0.1 Mass of NaOH formed = 0.1 \times 40 = 4 g Hence, correct answer is (B)

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