Answer:
Explanation:
to show how precise your measurements were.
<span>1.16 moles/liter
The equation for freezing point depression in an ideal solution is
ΔTF = KF * b * i
where
ΔTF = depression in freezing point, defined as TF (pure) ⒠TF (solution). So in this case ΔTF = 2.15
KF = cryoscopic constant of the solvent (given as 1.86 âc/m)
b = molality of solute
i = van 't Hoff factor (number of ions of solute produced per molecule of solute). For glucose, that will be 1.
Solving for b, we get
ΔTF = KF * b * i
ΔTF/KF = b * i
ΔTF/(KF*i) = b
And substuting known values.
ΔTF/(KF*i) = b
2.15âc/(1.86âc/m * 1) = b
2.15/(1.86 1/m) = b
1.155913978 m = b
So the molarity of the solution is 1.16 moles/liter to 3 significant figures.</span>
Answer:
ΔrxnHº =-1,124.3 kJ
Explanation:
To solve this question we first need to search for the enthalpies of formation of reactants and products and calculate the change in enthalpy of reaction utilizing the equation
ΔrxnHº = ∑ νΔfHº reactants - Σ νΔfHº products
where ν is the stoichiometric coefficient of the compound in the balanced equation .
ΔfHº H₂S(g) = -20.50 kJmol⁻¹
ΔfHº 0₂(g) = 0 ( O₂ is in standard state )
ΔfHº H₂O(l) = -285.8 kJmol⁻¹
ΔfHº SO₂(g) = -296.84 kJmol⁻¹
ΔrxnHº = [2 x -285.8 + 2 x -296.84] - [2 x -20.50]
= - 1,124.3 kJ