Answer:
127.3° C, (This is not a choice)
Explanation:
This is about the colligative property of boiling point.
ΔT = Kb . m . i
Where:
ΔT = T° boling of solution - T° boiling of pure solvent
Kb = Boiling constant
m = molal (mol/kg)
i = Van't Hoff factor (number of particles dissolved in solution)
Water is not a ionic compound, but we assume that i = 2
H₂O → H⁺ + OH⁻
T° boling of solution - 118.1°C = 0.52°C . m . 2
Mass of solvent = Solvent volume / Solvent density
Mass of solvent = 500 mL / 1.049g/mL → 476.6 g
Mol of water are mass / molar mass
76 g / 18g/m = 4.22 moles
These moles are in 476.6 g
Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m
T° boling of solution = 0.52°C . 8.85 m . 2 + 118.1°C = 127.3°C
Answer:
32.8g/mole
Explanation:
Given parameters:
Mass of sample of gas = 32.8g
Volume = 22.4L
Unknown:
Molecular weight = ?
Solution:
To solve this problem we must understand that at rtp;
1 mole of gas occupies a volume of 22.4L
Number of mole of the gas = 1 mole
Now;
Mass = number of moles x molecular weight
molecular weight = = = 32.8g/mole
Explanation:
Whether you choose to use van der Waals radii or metallic radii as a measure of the atomic radius, for metals the ionic radius is smaller than either, so the problem doesn't exist to the same extent. It is true that the ionic radius of a metal is less than its atomic radius (however vague you are about defining this).
Answer:
solubility of X in water at 17.0 is 0.11 g/mL.
Explanation:
Yes, the solubility of X in water at 17.0 can be calculated using the information given.
Let's assume solubility of X in water at 17.0 is y g/mL
The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.
So, solubility of X in 1 mL of water = y g
Hence, solubility of X in 36.0 mL of water = 36y g
So, 36y = 3.96
or, y = = 0.11
Hence solubility of X in water at 17.0 is 0.11 g/mL.