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The boiling point of benzene is 353.23 K...

The boiling point of benzene is 353.23 K. When 1.8 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to be 354.11 K. Calculate the molar mass of the solute, (`K_(b)` for benzene is 2.53 K kh `mol^(-1)`).

Text Solution

Verified by Experts

The correct Answer is:
57.5 kg
Molar mass of solution , `M_(2) = (W_(2)M_(1))/(W_(1)((p^(@) - P_(s))/(p^(@))))`
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The boiling point of benzene is 353.2 K. When 1.8 g of a non-volatile solute was dissolved in 90g benzene the boiling point was raised to 354.1 K. Calculate the molecular mass of the solute. ( K_(b) of benzene = 2.53 K kg mol^(-1) )

The boiling point of benzene is 353.2 K .When 1.8 g of a non - volatile solute was dissolved in 90 g benzene the boiling point was raised to 354.1 K . Calculate the molecular mass of the solute. ( K_b of benzene = 2.53 K kg mol^-1 )

Knowledge Check

  • The molal boiling point constant of water is 0.53^@C. When 2 mole of glucose are dissolved in 4000 g of water, the solution will boil at:

    A
    `100.53^@C`
    B
    `101.06^@C`
    C
    `100.265^@C`
    D
    `99.47^@C`

  • The molal boiling point constant of water is 0.53^@C. When 2 mole of glucose are dissolved in 4000 g of water, the solution will boil at:

    A
    `100.53^@C`
    B
    `101.06^@C`
    C
    `100.265^@C`
    D
    `99.47^@C`

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