SELECT * FROM question_mgmt as q WHERE id=1772 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=9 AND chapterId=57 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)
Calculate
(a) molality
(b) molarity and
(c) mole
fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
(a) Molar mass of KI = 39 + 127 = 166 g mol - 1
20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.
That is,
20 g of KI is present in (100 - 20) g of water = 80 g of water
Therefore, molality of the solution = Moles of KI / Mass of water in kg
= 20/166 / 0.08 m
= 1.506 m
= 1.51 m (approximately)
(b) It is given that the density of the solution = 1.202 g mL - 1
∴Volume of 100 g solution = Mass / Density
= 100g / 1.202g mL-1
= 83.19 mL
= 83.19 × 10 - 3 L
Therefore, molarity of the solution = 20/166 mol / 83.19 × 10 - 3 L
= 1.45 M
(c) Moles of KI = 20/166 = 0.12 mol
Moles of water = 80/18 = 4.44 mol
Therefore, mole fraction of KI = Moles of KI / (Moles of KI + Moles of water)
= 0.12 / (0.12+4.44)
= 0.0263
The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D
Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate of formation of D/mol L - 1 min - 1 |
I | 0.1 | 0.1 |
6.0 × 10 - 3 |
II | 0.3 | 0.2 |
7.2 × 10 - 2 |
III | 0.3 | 0.4 |
2.88 × 10 - 1 |
IV | 0.4 | 0.1 |
2.40 × 10 - 2 |
Determine the rate law and the rate constant for the reaction.
Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot ptotal' pchloroform' and pacetoneas a function of xacetone. The experimental data observed for different compositions of mixture is.
100 ×xacetone |
0 | 11.8 | 23.4 | 36.0 | 50.8 | 58.2 | 64.5 | 72.1 |
pacetone /mm Hg |
0 | 54.9 | 110.1 | 202.4 | 322.7 | 405.9 | 454.1 | 521.1 |
pchloroform/mm Hg |
632.8 | 548.1 | 469.4 | 359.7 | 257.7 | 193.6 | 161.2 | 120.7 |
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