Tutor profile: Anshul S.
Subject: Physical Chemistry
A 200 ml, 0.1 M Solution of K4[Fe(CN)6] completely reduces Iodine (I2) in acidic medium to KI. If the Iodine was formed through a reaction of acidified KMno4 with KI. Find the amount of KMno4 in grams required for the process.
K4[Fe(CN)6] will react with I2 to give Fe+3 ions, CO2, NO3- ions and KI. Now, the n-factor of K4[Fe(CN)6] will be sum of the n-factors of Fe, C and N. For Fe: Fe+2 --> Fe+3 + e- (n-factor = 1) For C: 6C+2 --> 6C+4 + 12e- (nfactor =12) For N 6N-3 --> 6N+5 + 48e- (nfactor = 48) Hence nfactor for K4[Fe(CN)6] = 1+12+48 = 61 Hence equivalents of K4[Fe(CN)6] = Molarity * Volume(L) * n-factor = 0.1*0.2*61 = 1.22 Now, using Law of equivalence we can say that the equivalents of I2 will also be 1.22 and hence the equivalents of KMn04 will also be 1.22. Now, KMno4 in acidic medium reduces to Mn+2 hence n-factor of KMno4 will be calculated as: Mn+7 + 5e ---> Mn+2 (n-factor =5) Hence Moles of KMno4 will be 1.22/5 =0.244 Hence mass of KMno4 will be 0.244*158= 38.55 g
Subject: Organic Chemistry
How many Diasteriomeric pairs are possible for 2-Bromo,3-Chloro,4-Fluro,5-Iodo Hexane?
Total number of Chiral Carbons in this compound is 4. Since the Compound does not have any symmetry, the number of stereoisomers will be simply 2^4 = 16. Total number of pairs of isomers that can be formed = 16C2 =120. There will be 8 pairs of Enantiomers out of the 120 pairs. Hence the remaining pairs will be Diasteriomeric pairs = 120-8 = 114.
A mixture of 1 mole each of Li2CO3, K2CO3 and Na2CO3 was heated at 200 degrees Celsius. The liberated gas was completely passed through excess of lime water solution resulting in formation of a white precipitate. What mass of the precipitate will form?
Since Na2CO3 and K2CO3 do not decompose at 200 Degree Celsius, Only Li2CO3 will decompose through the following reaction: Li2CO3 ---> Li2O + CO2 Hence 1 mole of Li2CO3 will give out 1 mole of CO2. 1 Mole of CO2 will react with excess of Lime water Ca(OH)2 through the following reaction: CO2 + Ca(OH)2 ---> CaCO3 + H2O Hence the white precipitate is of CaCO3. 1 Mole of CO2 will give 1 mole of CaCO3. Mass of 1 mole of CaCO3 = 100g Hence 100 g of the white precipitate will form.
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