Question

unatempted.
A 0.200 g sample containing Cu(II) was analyzed iodometrically :
Cu(II) is reduced to copper(I) by iodine
2Cu2+ +41-2Cul +1,.
If 20 ml 0.10 M Na,s,o, is required for titration of liberated I,
then the percentage of copper in sample will be (Cu = 63.5 ; I =
127)
(A) 31.75%
(C) 53%
(B) 63,5%
(D) 37%​

Answer 1

Answer:

Correct answer is option 'B'. Can you explain this answer?

Related Test: Part Test 1 - NEET 2020

NEET Question

By Vaibhav Chaudhary · Aug 30, 2020 ·NEET

Answer ReportShareSave

Kartik Bansal answered Feb 12, 2019

2 moles of Cu2+ = 1 mole of I2

= 2 moles of hypo.

so moles of hypo used = 20 x 10-3 x 0.1 = 2 milli moles = milli moles of copper

Hence percentage of copper

Upvote

This discussion on A 0.200 gm sample containing copper (II) was analysed iodometrically, copper (II) is reduced tocopper (I) by iodine. 2Cu2+ + 4I–——? 2 CuI + I2If 20.0mL of 0.10 M Na2S2O3 is required for titration of the liberated iodine then the percentage of copper in the sample will be (Cu = 63.5 g/mole)a)31.75 %b)63.5 %c)53 %d)37 %Correct answer is option 'B'. Can you explain this answer? is done on EduRev Study Group by NEET Students. The Questions and Answers of A 0.200 gm sample containing copper (II) was analysed iodometrically, copper (II) is reduced tocopper (I) by iodine. 2Cu2+ + 4I–——? 2 CuI + I2If 20.0mL of 0.10 M Na2S2O3 is required for titration of the liberated iodine then the percentage of copper in the sample will be (Cu = 63.5 g/mole)a)31.75 %b)63.5 %c)53 %d)37 %Correct answer is option 'B'. Can you explain this answer? are solved by group of students and teacher of NEET, which is also the largest student community of NEET. If the answer is not available please wait for a while and a community member will probably answer this soon. You can study other questions, MCQs, videos and tests for NEET on EduRev and even discuss your questions like A 0.200 gm sample containing copper (II) was analysed iodometrically, copper (II) is reduced tocopper (I) by iodine. 2Cu2+ + 4I–——? 2 CuI + I2If 20.0mL of 0.10 M Na2S2O3 is required for titration of the liberated iodine then the percentage of copper in the sample will be (Cu = 63.5 g/mole)a)31.75 %b)63.5 %c)53 %d)37 %Correct answer is option 'B'. Can you explain this answer? over here on EduRev! Apart from being the largest NEET community, EduRev has the largest solved Question bank for NEET.