Question

A certain liquid X has a normal boiling point of 118.90 °C and a boiling point elevation constant Kb = 0.82 °C*kg*mol^-1. Calculate the boiling point of a solution made of 54.g of potassium bromide (KBr) dissolved in 750. g of X. Be sure your answer is rounded to the correct number of significant digits.

Answer 1

Answer: The boiling point of solution is
`1.2* 10^2^oC`

Step-by-step explanation:

To calculate the elevation in boiling point, we use the equation:

`\Delta T_b=iK_bm`

Or,

`\Delta T_b=i* K_b* \frac{m_(solute)* 1000}{M_(solute)* W_(solvent)\text{ in grams}}`

where,

`\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}`

Boiling point of pure liquid = 118.90°C

i = Vant hoff factor = 2 (For potassium bromide)

`K_b` = molal boiling point elevation constant = 0.82°C/m

`m_(solute)` = Given mass of solute (potassium bromide) = 54. g

`M_(solute)` = Molar mass of solute (potassium bromide) = 119 g/mol

`W_(solvent)` = Mass of solvent (liquid X) = 750. g

Putting values in above equation, we get:

`\text{Boiling point of solution}-118.90=2* 0.82^oC/m* (54* 1000)/(119g/mol* 750)\\\\\text{Boiling point of solution}=119.9^oC=1.2* 10^2^oC`

Hence, the boiling point of solution is
`1.2* 10^2^oC`