1.

The boiling pointof benzene is353 .23 K. When1.80gramof non- volatile solutewasdissolvedin 90gramof benzene , theboilingpoint is raisedto 354.11 K. Calculatethe molarmass ofsolute . [K_(b)for benzene = 2.53 K kg mol^(-1)]

Answer»


Solution :Given : `T_(b)^(o) =353 K , T_(b)= 354 . 11 K`
`W_(1) = 90 g , W_(2)= 1.8 , K_(b)= 2.53 K kg mol^(-1)`
Molar mass`=M_(2) = ?`
`DELTA T_(b) = T_(b)- T_(b)^(o) = 354.11 - 353 .23 = 0.88K`
`Delta T_(b) = K_(b) XX (W_(2) xx 1000)/(W_(1) xx M_(2))`
`:. M_(2)= (K_(b) xx W_(2) xx 1000)/(Delta T_(b)xx W_(1))`
` = (2.53 xx1.8 xx 1000)/(0.88 xx90)`
` =57.5 g "mol"^(-1)`


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