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Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water

Answer»

Solution :The DATA provided is WRITTEN as under: `p^0 = 17.535 mm, w_2 = 25g, w_1 = 450 g `
For solute (glucose `C_6H_12O_6`) , `M_2 = 180 g "mol"^(-1)`
For solvent ` (H_2O) , M_1 = 18g "mol"^(-1)`
Applying Raoult.s LAW,
`(p^0 - p_s)/(p^0) = (n_2)/(n_1+ n_2) " or" (p^0 -p_s)/(p_s) =n_2/n_1= (w_2// M_2)/(w_1//M_1)" or " (p^0)/(p_s)- 1 = (w_2M_1)/(w_1 M_2)`
Substituting the values in the above equation, we get
`(17.535)/(p_s) - 1 = (25 xx 18)/(450 xx 180) = 25/4500 " or" (17.535)/(p_s) = 1+ 25/4500 = 4525/4500`
`p_s =17.535 xx 4500/4525 = 17.44 mm`


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