A solution containing 0.73 g of camphor ( molar mass 152 g `"mol"^(-1)` ) in 36.8 g of acetone ( boiling point `56.3^(@)C)` boils at `56* 55^(@)C` . A solution of 0.564 g of unknown compound in the same weight of acetone boils at `56.46^(@)C` . Calculate the molar mass of the unknown compound .

Explanation

Given : Mass of camphor = `W_(2) = 0*73` g
Molar mass of camphor = `M_(2) = 152 g "mol"^(-1)`
Mass of acetone = `W_(1) = 36*8 ` g
Mass of unknown compound = ` W'_(2) = 0*564` g
Molar mass of compound = `M'_(2)= ?`
For solution of camphor in acetone ,
` DeltaT_(b) = T_(b)- T^(@)._(b)`
` = 329 * 55 - 329*30 = 0*25 ` K
` :. K_(b) = (Delta T_(b)xx W_(1)xxM_(2))/(W_(2) xx 1000)`
` = (0*25 xx 36*8 xx 152)/(0*73 xx 1000)`
` = 1*916 "K kg mol"^(-1)`
For solution of unknown compound in acetone,
` DeltaT_(b) = T_(b) - T^(@)._(b)`
` = 329*46 - 329*30 = 0*16 ` K
` :. K_(b) = (DeltaT_(b)xxW_(1)xxM'_(2))/(E'_(2)xx1000)`
` M'_(2) = (K_(b)xxW'_(2)xx1000)/(DeltaT_(b)xxW_(2))`
` = (1*916 xx1000 xx0.564)/(0*16 xx36*8)`
` M'_(2) = 185*5` g/mol
Hence , the molar mass of the unknown compound is ` 183*5 "g mol"^(-1)`