Question

Vapour pressure of pure water is 760mm at 25°C. The vapour pressure of solution containing 1m solution of glucose will be ?

Answer 1

1 answer · Chemistry 

 Best Answer

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) 

Answer 2

Answer : The vapor pressure of solution is,  746.32 mm Hg

Explanation:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

$\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}$

where,

$p^o$ = vapor pressure of pure solvent  (water) = 760 mm Hg

$p_s$ = vapor pressure of solution  = ?

Given : Molality of glucose solution = 1m = 1 mole is dissolved in 1 kg of solvent

$w_2$ = mass of solute  (glucose)=$moles\times {\text {Molar mass}}=1\times 180g/mol=180g$

$w_1$ = mass of solvent  = 1 kg = 1000 g

$M_1$ = molar mass of solvent (water) = 18 g/mole

$M_2$ = molar mass of solute (glucose)= 180 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

$\frac{760-p_s}{760}=\frac{180\times 18}{1000\times 180}$

$p_s=746.32 mmHg$