Question

Calculate the boiling point of a solution containing 25 g urea and 25 g thiourea in 500 g chloroform. the boling point of pure chloroform is 61.2°C and Kb = 3.63 K m-¹​

Answer 1

Answer :

• Boiling point of solution = 66.616°C.

Step-by-step explanation :

Given that,

• Boiling point of pure chloroform = 61.2°C

• $\bold{K_b=3.63\:K\:m^{-1}}$

$\hrulefill$

No. of moles of urea =

$\bold{\dfrac{25}{60}mole=0.417\:mole}$

No. of moles of thiourea =

$\bold{\dfrac{25}{76}mole=0.329\:mole}$

Total moles = 0.417 + 0.329 = 0.746

Mass of solvent (chloroform) = 500 g = 0.500 kg

∴ $\bold{Molality=\dfrac{Moles\:of\:solute}{Mass\:of\:solvent\:in\:kg}}$

$\implies\bold{Molality=\dfrac{0.746\:mol}{0.500\:kg}}$

$\implies\bold{Molality=1.492\:m}$

$\rule{200}{3}$

Now, $\bold{\Delta\:T_b=K_b\times{}m}$

$\bold{\Delta\:T_b=3.63\times{}1.492=5.416^{\circ}}$

∴ Boiling point of solution = 61.2 + 5.416°C = 66.616°C

Answer 2

Answer:

solve in this way. ...................