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)

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 

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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 

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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)