19. When 0.5g of calcium trioxocarbonate (IV) was added to excess dilute
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This problem is incomplete. Luckily, I found a similar problem from another website shown in the attached picture. The data given can be made to use through the Clausius-Clapeyron equation:
ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁)
where
P₁ = 14 Torr * 101325 Pa/760 torr = 1866.51 Pa
T₁ = 345 K
P₂ = 567 Torr * 101325 Pa/760 torr = 75593.78 Pa
T₂ = 441 K
ln(75593.78 Pa/1866.51 Pa) = (-ΔHvap/8.314 J/mol·K)(1/441 K - 1/345 K)
Solving for ΔHvap,
<em>ΔHvap = 48769.82 Pa/mol or 48.77 kPa/mol</em>
ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁)
where
P₁ = 14 Torr * 101325 Pa/760 torr = 1866.51 Pa
T₁ = 345 K
P₂ = 567 Torr * 101325 Pa/760 torr = 75593.78 Pa
T₂ = 441 K
ln(75593.78 Pa/1866.51 Pa) = (-ΔHvap/8.314 J/mol·K)(1/441 K - 1/345 K)
Solving for ΔHvap,
<em>ΔHvap = 48769.82 Pa/mol or 48.77 kPa/mol</em>