Find the boiling point of a solution of 2.00 m solution of sodium chloride, NaCl, in water (kb= 0.512°C, bp= 100.0°C)

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Find the boiling point of a solution of 2.00 m solution of sodium chloride, NaCl, in water (kb= 0.512°C, bp= 100.0°C)

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Cherry 6 months 2021-07-21T02:07:24+00:00 1 Answers 13 views 0

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    2021-07-21T02:09:22+00:00

    Answer: Boiling point of the given solution is 102.048^{o}C.

    Explanation:

    Given: Molality = 2.00 m

    k_{b} = 0.512^{o}C

    Now, equation for dissociation of water is as follows.

    H_{2}O \rightarrow H^{+} + OH^{-}

    As it is giving 2 ions upon dissociation. So, the value of i = 2.

    Formula used to calculate change in temperature is as follows.

    \Delta T = i \times k_{b} \times m

    where,

    i = Van’t Hoff factor

    k_{b} = molal boiling point elevation constant

    m = molality

    Substitute the values into above formula as follows.

    \Delta T = i \times k_{b} \times m\\= 2 \times 0.512^{o}C \times 2.00 m\\= 2.048^{o}C

    As the boiling point of water is 100^{o}C. Hence, the boiling point of solution will be as follows.

    \Delta T^{'}_{b} = 100^{o}C + 2.048^{o}C\\= 102.048^{o}C

    Thus, we can conclude that boiling point of the given solution is 102.048^{o}C.

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