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) 37.04% b) 37.04% c) 9.63%
Explanation:
The theoretical percent recovery (Tr), is the total percentage of each compound in the sample. Depending on the technique used to recovery the compounds, the percent recovery will be less than the theoretical, because no technique is 100% efficient.
So, to calculate the theoretical, it will be the mass of the compound divided by the mass of the sample multiplied by 100%.
a) Tr = (250 mg)/(675 mg) * 100%
Tr = 37.04%
b) Tr = (250 mg)/(675 mg) * 100%
Tr = 37.04%
c) Tr = (65 mg)/(675 mg) * 100%
Tr = 9.63%
When cohesive force is stronger than the adhesive force: concave up meniscus, water forms droplets on surface
Answer:
17,932.69 g/mol is the molecular weight of the substance.
Explanation:
Using Beer-Lambert's law :
Formula used :
where,
A = absorbance of solution = 1.04
c = concentration of solution =?
l = length of the cell = 1 cm
= molar absorptivity of this solution = 18,650 
Now put all the given values in the above formula, we get the molar absorptivity of this solution.
c = 
V = Volume of the solution in L
Molecular weight of the substance = x
V = 100 mL = 0.1 L
Mass of the substance = 100 mg = 0.1 g
x = 17,932.69 g/mol
17,932.69 g/mol is the molecular weight of the substance.
I'm not exactly sure which one but I do know that an acid and a base react in a aqueous solution to form water, so i would probably eliminate the ones that aren't aqueous solutions.
