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:
H2O is a compound because its a main constitute of earths hydrosphere
Answer:
1.62 L
Explanation:
T= 97+273.15= 370.15
R= 0.08206 atm/mol⋅K
V= 45 L
n= 2.4 mol
P= (n⋅R⋅T)/V
= (2.4 x 0.08206 x 370.15)/(45) = 1.61997 = 1.62
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, 
molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in 
of ethane gas
And, as we know that
1 mole of ethane molecule contains 
molecules of ethane
2.869 moles of ethane molecule contains 
molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
