The boiling point of benzene is 353 .23 K. When 1.80 gram of non- volatile solute was dissolved in 90 gram of benzene , the boiling point is raised to
The boiling point of benzene is 353 .23 K. When 1.80 gram of non- volatile solute was dissolved in 90 gram of benzene , the boiling point is raised to 354.11 K. Calculate the molar mass of solute .
`[K_(b)`for benzene = 2.53 K kg `mol^(-1)]`
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Correct Answer - Molar mass `=M_(2) =57.5 g mol^(-1)`
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 xx 1.8 xx 1000)/(0.88 xx 90)`
` =57.5 g "mol"^(-1)`
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