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The boiling a point of benzene is 353.23...

The boiling a point of benzene is 353.23K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. `K_(b)` for benzene is 2.53 K kg `mol^(-1)`.

A

`5.8 g mol^(-1)`

B

`0.58 g mol^(-1)`

C

`58 g mol^(-1)`

D

`0.88 g mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
The elevation `(Delta T_(b))` in the boiling point `= 354.11 K - 353.23 K = 0.88 K`
Substituting these values in expression
`M_("Solute")=(K_(b)xx1000xxw)/(Delta T_(b)xx W)`
Where, w = weight of solute, W = weight of solvent
`M_("solute")=(2.53xx1.8xx1000)/(0.88xx90)=58 gm mol^(-1)`
Hence, molar mass of the solute `= 58 gm mol^(-1)`
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The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_(b) for benzene is 2.53 K kg mol^(-1) .

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. (K_b " for benzene is " 2.53 K kg mol^(-1))

Knowledge Check

  • The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_b , for benzene is 2.53 K kg mol^(-1) .

    A
    `58 g mol^(-1)`
    B
    `85 g mol^(-1)`
    C
    `93 g mol^(-1)`
    D
    `108 g mol^(-1)`

  • When 2.0g of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point of benzene is raised by 0.88K. Which of the following may be the solute? K_(b) for benzene =2.53 K kg mol^(-1))

    A
    `CO(NH_(2))_(2)`
    B
    `C_(6)H_(12)O_(6)`
    C
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    D
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