Answer:
130.4 grams of sucrose, would be needed to dissolve in 500 g of water.
Explanation:
Colligative property of boiling point elevation:
ΔT = Kb . m . i
In this case, i = 1 (sucrose is non electrolytic)
ΔT = Kb . m
0.39°C = 0.512°C/m . m
0.39°C /0.512 m/°C = m
0.762 m (molality means that this moles, are in 1kg of solvent)
If in 1kg of solvent, we have 0.712 moles of sucrose, in 500 g, which is the half, we should have, the hallf of moles, 0.381 moles
Molar mass sucrose = 342.30 g/m
Molar mass . moles = mass
342.30 g/m . 0.381 m = 130.4 g
Answer:
a. True
Explanation:
The data curve for OCI-/H3O appears to be similar to C2H3O2-/H3O because protons have not dissociated with conjugate base. This conjugate base is strong and hold proton in solution while acids try to dissociate it.
