1. Tardigrade
  2. Question
  3. Chemistry
  4. The boiling point of pure benzene is 80.1° C. When 2 g of a non-volatile solute (molar mass =60 g mol -1 ) was dissolved in 100 g of benzene. The solution will boil at ° C. ( K b . for benzene .=2.53 K m -1)

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Q. The boiling point of pure benzene is $80.1^{\circ} C$. When $2\, g$ of a non-volatile solute (molar mass $=60\, g\, mol ^{-1}$ ) was dissolved in $100\, g$ of benzene. The solution will boil at_____ ${ }^{\circ} C$.
$\left( K _{ b }\right.$ for benzene $\left.=2.53\, K\, m ^{-1}\right)$

Solutions

Solution:

$\Delta T _{ b }= K _{ b } \frac{ W _{ B }}{ M _{ B } W _{ A }}$
$=2.53 \times \frac{2}{60 \times 0.1}$
$=0.84\, K$
or $0.84^{\circ} C$
$\Delta T _{ b }= T _{ b }- T _{ b }^{\circ}$
$\therefore T _{ b }= T _{ b }^{\circ}+\Delta T _{ b }$
$=80.1+0.84$
$=80.94{ }^{\circ} C$