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Chemical calculations with explanatory notes, problems, and answers, specially adapted for use in colleges and science schools - Softcover
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1896 edition. Excerpt: ...are required. Calculate the percentage of chlorine in the specimen. The equations representing the reactions are--KC103 + 3H, = KC1 + 3H20; KC1 + AgN03 = AgCl + KN03. Since the mol. wgt. of AgN03 = 170, a normal solution will contain 170 grams in 1 litre of solution;. 1 c.c. will contain 0-170 gram AgN03. If the solution is deci-normal, since only--of 170 J 10 grams is dissolved, 1 c.c. will equal 0-0170 gram AgN03. Now 170 grams AgN03 will precipitate 35-5 grams of chlorine as AgCl;.-. 1 c.c. of N AgN03 (= 0-17 gram) will precipitate (or is equivalent to) 0-0355 gram of chlorine; and.'. 1 c.c. of-AgN03 (= 0-017 gram) will precipitate (or is equivalent to) 0 00355 gram of chlorine. N In the example given, 26 c.c. of--AgN03 were used;, 26 x 0-00355 = the weight of 01, to which they are equivalent; and, since that is obtained from 0-32 gram KC103, the percentage amount is found by the following proportion:--0-32: 100:: 26x0-00355: x; 100 x26 x0-00355 oa a... x =----= 26-64: per cent, chlorine. L)-o-Z-Example 6.--100 c.c. of a solution of sulphurous acid re-N quired 5-15 c.c. of--iodine solution (1 c.c. = 0-0127 gm.) to completely oxidise it. Find the weight of SfX per litre of solution. S02 + I2 + 2H20 = H2S04 + 2HI. 1 To titrate a solution means to ascertain the titre, or strength, of it by volumetric methods. Since one molecule of iodine oxidises one molecule of sulphur dioxide; therefore--254 parts of iodine oxidise 64 parts of sulphur dioxide; now 100 c.c. of solution required 5-15 x 0-0127 gm. iodine;. lOOOc.c. „ „ require 51-5 x0-0127 „ „ and.-. 254: 51-5 x 0-0127:: 64: x, the S02 in solution. 51-5 x 0-0127 x 64 ntc-Aa nn--, x =---= 0 1648 gm. S02 in 1 litre. 254 Questions. 1. 1-2277 grams of crystallised copper...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1896 edition. Excerpt: ...are required. Calculate the percentage of chlorine in the specimen. The equations representing the reactions are--KC103 + 3H, = KC1 + 3H20; KC1 + AgN03 = AgCl + KN03. Since the mol. wgt. of AgN03 = 170, a normal solution will contain 170 grams in 1 litre of solution;. 1 c.c. will contain 0-170 gram AgN03. If the solution is deci-normal, since only--of 170 J 10 grams is dissolved, 1 c.c. will equal 0-0170 gram AgN03. Now 170 grams AgN03 will precipitate 35-5 grams of chlorine as AgCl;.-. 1 c.c. of N AgN03 (= 0-17 gram) will precipitate (or is equivalent to) 0-0355 gram of chlorine; and.'. 1 c.c. of-AgN03 (= 0-017 gram) will precipitate (or is equivalent to) 0 00355 gram of chlorine. N In the example given, 26 c.c. of--AgN03 were used;, 26 x 0-00355 = the weight of 01, to which they are equivalent; and, since that is obtained from 0-32 gram KC103, the percentage amount is found by the following proportion:--0-32: 100:: 26x0-00355: x; 100 x26 x0-00355 oa a... x =----= 26-64: per cent, chlorine. L)-o-Z-Example 6.--100 c.c. of a solution of sulphurous acid re-N quired 5-15 c.c. of--iodine solution (1 c.c. = 0-0127 gm.) to completely oxidise it. Find the weight of SfX per litre of solution. S02 + I2 + 2H20 = H2S04 + 2HI. 1 To titrate a solution means to ascertain the titre, or strength, of it by volumetric methods. Since one molecule of iodine oxidises one molecule of sulphur dioxide; therefore--254 parts of iodine oxidise 64 parts of sulphur dioxide; now 100 c.c. of solution required 5-15 x 0-0127 gm. iodine;. lOOOc.c. „ „ require 51-5 x0-0127 „ „ and.-. 254: 51-5 x 0-0127:: 64: x, the S02 in solution. 51-5 x 0-0127 x 64 ntc-Aa nn--, x =---= 0 1648 gm. S02 in 1 litre. 254 Questions. 1. 1-2277 grams of crystallised copper...
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