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Grade: 12
        
The vapour pressure of pure water is 760 mm at 25 degree celsius. The vapour pressure of solution containing 1(m) solution of glucose will be?? (Please solve using Raoult’s Law)
5 months ago

Answers : (1)

Arun
22967 Points
							
use raoult's law for this 
.. PA = (P*A) x (χA) 
where 
.. PA = vapor pressure of component A over the liquid 
.. P*A = vapor pressure of pure component A 
.. χA = mole fraction of component A in the liquid phase 
****** 
the key here is for you to recognize that glucose is non-volatile. So that the pressure over the solution will = the vapor pressure of WATER in that solution. 
i.e.. that "A" in the above equation = H2O.. and Ptotal = PA 
AND... you need to convert from 1m glucose to mole fraction H2O 
********** 
can YOU finish? 
hints....on that conversion. 
.. (1) use dimensional analysis 
.. (2) 1m means 1 mole glucose / 1kg H2O 
.. (3) assume 1kg H2O, convert that to moles H2O and moles glucose 
.. (4) mole fraction H2O = mole H2O / (mole glucose + mole H2O) 
.. (5) your setup should look like this 
.. .. . ..1kg H2O... 1mol glucose 
.. .. ... --- --- ----- x -- --- --- --- ---- = __ mol glucose 
.. .. .. .. .. ..1.. .. .. .. . .1kg H2O 
.. ... .. . 1kg H2O.. .. .1000g... .. .1 mol 
.. ... ... ---- ----- --- x ---- ---- --- x --- ---- --- = __mol H2O 
... ... ... .. ...1... ... .. ... 1kg.. ..... .18.02g 
.. (5) Ptotal = (P*A) x (χA) 
 
5 months ago
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