Answer:
283.725 kJ ⋅ mol − 1
Explanation:
C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1
Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1
C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1
4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1
eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1
so,
average bond enthalpy is = 283.725 kJ ⋅ mol − 1
Answer : The molecular weight of a substance is 157.3 g/mol
Explanation :
As we are given that 7 % by weight that means 7 grams of solute present in 100 grams of solution.
Mass of solute = 7 g
Mass of solution = 100 g
Mass of solvent = 100 - 7 = 93 g
Formula used :
where,
= change in freezing point
= temperature of pure water =
= temperature of solution =
= freezing point constant of water =
m = molality
Now put all the given values in this formula, we get
Therefore, the molecular weight of a substance is 157.3 g/mol