Question

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)

Answer 1

Answer: Boiling point of the given solution is
`102.048^(o)C`.

Step-by-step 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 * k_(b) * 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 * k_(b) * m\\= 2 * 0.512^(o)C * 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`.