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What would be the molar mass of compound if 6.21 g of it is dissolved in 24.0 g of CH Cl_(3) from a solution that has a boiling point of 68.04^(@)C. The boiling point of pure chloroform is 61.7^(@)C and the boiling point elevation constant Kb for chloroform is 3.63^(@)C/m. |
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Answer» Solution :Elevation in boiling point `DeltaT_(b) = 68.04 - 61.7 = 6.31 ^(@) C` Mass of substance `W_(B) = 6.21` g Mass of `CH Cl_(3) W_(A) = 24.0` g `K_B = 3.63 ""^(@) C//m` `M_(B) = (K_(b) xx W_(B) xx 1000)/(DeltaT_(b) xx W_(A)) = (3.63 xx 6.21 xx 1000)/(6.34 xx 24)` |
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