how much of cane sugar should be dissolved in 0.300Kg so that resulting solution has same boiling point as that of 3gm of Urea in 0.1Kg of Water
Asked by Hritwik mishra | 19th Aug, 2011, 07:13: PM
The elevation in boiling point for sugar solution is,
?Tb(sucrose) = (Kb X Wsucrose) / (Msucrose X Wwater )
Where, Msucrose = 342 gm/mole
Wwater = 0.300 kg
Similarly, the elevation in boiling point for urea solution is,
?Tb(urea) = (Kb X Wurea ) / (Murea X Wwater )
Where, Murea = 60 gm/mole
Wwater = 0.1 kg
According to the question, ?Tb(sucrose) = ?Tb(urea)
(Kb X Wsucrose) / (Msucrose X Wwater ) = (Kb X Wurea ) / (Murea X Wwater )
Or, Wsucrose / (342 X 0.300 ) = 3 / ( 60 X 0.1 )
Wsucrose = 51.33 gm
The elevation in boiling point for sugar solution is,
?Tb(sucrose) = (Kb X Wsucrose) / (Msucrose X Wwater )
Where, Msucrose = 342 gm/mole
Wwater = 0.300 kg
Similarly, the elevation in boiling point for urea solution is,
?Tb(urea) = (Kb X Wurea ) / (Murea X Wwater )
Where, Murea = 60 gm/mole
Wwater = 0.1 kg
According to the question, ?Tb(sucrose) = ?Tb(urea)
(Kb X Wsucrose) / (Msucrose X Wwater ) = (Kb X Wurea ) / (Murea X Wwater )
Or, Wsucrose / (342 X 0.300 ) = 3 / ( 60 X 0.1 )
Wsucrose = 51.33 gm
Answered by | 23rd Aug, 2011, 04:02: PM
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