Two elements A and B form compounds having formula AB 2 and AB 4 . When dissolved in 20 g of benzene, 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB 4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1K kg mol -1 . Calculate atomic masses of A and B.

Question

Philoid · Accepted Answer

let the molar masses of AB₂ and AB₄ be x and y respectively.

Molar mass of benzene(C₆H₆) = 12 × 6 + 1 × 6 = 78g/mol

Moles of benzene = mass/molar mass = 20/78

n = 0.256mol

⇒ ΔT_f = 2.3k

K_f = 5.1K kg mol^-1

For AB₂

Applying the formula: ΔT_f = K_f × M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ 2.3 = 5.1 × M₁

⇒ M₁ = 0.451mol/kg

For AB₄

Applying the formula: ΔT_f = K_f × M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ 1.3 = 5.1 × M₂

⇒ M₂ = 0.255mol/kg

M₁ = moles of solute/mass of solvent(in kg)

⇒

⇒ X = 110.86g

M₂ = moles of solute/mass of solvent(in kg)

⇒

⇒ y = 196.1g

atomic mass of A be a and that of B be b g respectively.

So, AB₂ : a + 2b = 110.86

AB_{4 :} a + 4b = 196.1

⇒ a = 25.59g

⇒ b = 42.64g

Two elements A and B form compounds having formula AB2 and AB4 . When dissolved in 20 g of benzene, 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1K kg mol-1. Calculate atomic masses of A and B.

Two elements A and B form compounds having formula AB₂ and AB₄ . When dissolved in 20 g of benzene, 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1K kg mol^-1. Calculate atomic masses of A and B.