<span>the behavior of the above pair of substances</span> is soluble
Explanation:
The given data is as follows.
Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml
- Therefore, molarity of toulene is calculated as follows.
Molarity =
=
= 4.11 M
Hence, molarity of toulene is 4.11 M.
- As molality is the number of moles of solute present in kg of solvent.
So, we will calculate the molality of toulene as follows.
Molality =
=
= 8.6 m
Hence, molality of given toulene solution is 8.6 m.
- Now, calculate the number of moles of toulene as follows.
No. of moles =
=
= 0.8227 mol
Now, no. of moles of benzene will be as follows.
No. of moles =
=
= 1.2239 mol
Hence, the mole fraction of toulene is as follows.
Mole fraction =
=
= 0.402
Hence, mole fraction of toulene is 0.402.
- As density of given solution is 0.857 so, we will calculate the mass of solution as follows.
Density =
0.857 = (As 1 = 1 g)
mass = 171.4 g
Therefore, calculate the mass percent of toulene as follows.
Mass % =
=
= 44.22%
Therefore, mass percent of toulene is 44.22%.
Answer:
Hello
The answer is 18,7(b)
Explanation:we don't know the value of o² so at first we should put (x)instead of gr of o²and then write 1molO²/32grO²×2mol SO²/3mol O²×64gr SO²/1molSO²=25 gr SO².and then just find the value of (x).
Answer:
One of the errors for low percentage of magnesium could be because not all the magnesium may have reacted.
Explanation:
During the heating process, if the magnesium have not reacted completely, it can lead to low percentage of magnesium in the oxide formed. The product may still look a bit greyish rather than whitish after the heating process.
The balanced equation for the reaction between Mg and HCl is as follows
Mg + 2HCl --> MgCl₂ + H₂
stoichiometry of HCl to H₂ is 2:1
number of HCl moles reacted - 0.400 mol/L x 0.100 L = 0.04 mol of HCl
since Mg is in excess HCl is the limiting reactant
number of H₂ moles formed - 0.04/2 = 0.02 mol of H₂
we can use ideal gas law equation to find the volume of H₂
PV = nRT
where
P - pressure - 1 atm x 101 325 Pa/atm = 101 325 Pa
V - volume
n - number of moles - 0.02 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 0 °C + 273 = 273 K
substituting these values in the equation
101 325 Pa x V = 0.02 mol x 8.314 Jmol⁻¹K⁻¹ x 273 K
V = 448 x 10⁻⁶ m³
V = 448 mL
therefore answer is
c. 448 mL