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`        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

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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