1.

Inan experiment , 1804 gof mannitolweredissolvedin 100 g of water . The vapour pressureof waterwaslowered by 0.309 mm Hgfrom 17.535 mm Hg. Calculate the molarmass ofmannitol .

Answer»


Solution :Given : Mass of a solute (MANNITOL) =`W_(2) =18.04 g`
Mass of a solvent(water) `=W_(1) =100g`
Vapourpressureof a solvent (water) `=P_(o)=17.535 mm HG`
Lowering ofvapourpressure `=Delta P= 0.309 g mm Hg`
Molarmass ofof `H_(2) O = M_(1) =18 g mol^(-1)`
Molarmass ofsolute (mannitol) = ?
`(P_(o)-P)/(P_(o)) = (W_(2) XX M_(1))/(W_(1) xx M_(2))`
`:' P_(o)- P= DeltaP`
`(DeltaP)/(P_(o)) = (W_(2) xx M_(1))/(W_(2) xx M_(2))`
`:. M_(2) = (P_(o))/(DeltaP) xx (W_(2) xx M_(1))/(W_(1))`
`= (17. 535 xx 18.04 xx 18)/(0.309 xx 100)`
`= 184 . 3 g "mol"^(-1)`


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