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

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phucdien · Answer 1 · 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 $102.048^{o}C$ .