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 $1.2\times 10^2^oC$

Explanation:

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

$\Delta T_b=iK_bm$

Or,

$\Delta T_b=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

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:

$\text{Boiling point of solution}-118.90=2\times 0.82^oC/m\times \frac{54\times 1000}{119g/mol\times 750}\\\\\text{Boiling point of solution}=119.9^oC=1.2\times 10^2^oC$

Hence, the boiling point of solution is $1.2\times 10^2^oC$