the boiling point of pure water is 100 degree Celsius calculate the boiling point of an aqua solution containing 0.6 gram of Urea( molar mass =60 g)in hundred gram of water( Kb for water is 0.52 K/m)
Answers
Mass of solvent (WA) = 100g
Mass of solute (WB) = 18g
Molecular mass of solute (MB) = 180
Molarity =
∆Tb = kb x m = 0.52x1 = 0.52
Boiling point of solution = [Boiling point of pure solvent] + [Elevation in boiling point]
= 373 + 0.52 = 373.52K
Answered by | 4th Jun, 2014, 03:23: PM
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Answer:
Mass of solvent (WA) = 100g
Mass of solute (WB) = 18g
Molecular mass of solute (MB) = 180
Molarity =
∆Tb = kb x m = 0.52x1 = 0.52
Boiling point of solution = [Boiling point of pure solvent] + [Elevation in boiling point]
= 373 + 0.52 = 373.52K
Answered by | 4th Jun, 2014, 03:23: PM
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Learn the meaning of elevation in boiling point & depressio...
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12gram glucose ko 10 gm solution Ka kwathnank 100.34 c hai,molal unnayan sthirank batao
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Calculate the boiling point of a solution containing 0.45g of camphor (mol. wt. 152) dissolved in 35.4g of acetone (b.p. 56.3°C); Kb per 100 gm of acetone is 17.2°C. IN THIS QUESTION WE USE THE FORMULA = 100 Kb *w2/w1M2 I want to know how we derive this formula in which 100 is in numerator and not 1000 . I know Kb is per 100 g though I am not able to derive this relation
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News18 India
CricketNext
Bangla News
Gujarati News
Urdu News
Marathi News
Moneycontrol
Firstpost
CompareIndia
History India
MTV India
In.com
Burrp
E-Learning Franchise Opportunity
Explore
Copyright Notice © 2019 Greycells18 Media Limited and its licensors. All rights reserved.