The mole fraction of ethanol c2h5oh in an aqueous solution that is 46 percent ethanol by mass is 0.25
Mole fraction is the mole of one of the component of the solution divided by the total moles of the solution
Calculating the mole fraction of ethanol,we get
Percentage by weight of ethyl alcohol is 46% which implies in 100 g of this solution is 46 g while the mass of the water will be 54% of 100 g = 54g.
Finding the moles of water and the ethyl alcohol {molar mass of water(H₂O) = 18 and molar mass of ethyl alcohol(C₂H₆O = 46}
Moles = mass/molar mass
moles of ethanol = 46/46 = 1
moles of water = 54/18 = 3
ntotal=1+3=4
Mole fraction of ethanol is:
Mole of ethanol = 1/4
Total mole of solution
Hence, the mole fraction is 1/4 or 0.25
To learn more about mole fraction visit:
brainly.com/question/8076655
#SPJ4
Was there supposed to be a picture?
<u>Answer:</u> The experimental van't Hoff factor is 1.21
<u>Explanation:</u>
The expression for the depression in freezing point is given as:
where,
i = van't Hoff factor = ?
= depression in freezing point = 0.225°C
= Cryoscopic constant = 1.86°C/m
m = molality of the solution = 0.100 m
Putting values in above equation, we get:
Hence, the experimental van't Hoff factor is 1.21
Answer:
Oxidation is the loss of electrons, that is, addition of electronegetive elements, example is addition of oxygen. Also, removal of electropositive elements, example is removal of hydrogen.
Explanation: a) In the presence of excess oxygen, propane burns in air, which gives the following chemical equation:
C3H8 + 5O2⇒ 3CO2 + 4H2O +Heat
b) When insufficient oxygen or too much oxygen is present for complete combustion, the following equation is given:
2C3H8 + 9O2 ⇒ 4CO2 + 2CO + 8H2O + Heat
c) At the anode( negative terminal): O∧2- ⇒ O + e
Oxygen accepts electron.
d) At cathode ( positive terminal): H∧+ + e∧- ⇒ H
Hydrogen donates electron
d) Nernst equation for reversal potential is given as follows:
E= RT/zF In{ion outside cell}/{ion inside cell}= 2.303 RT/zF In{ion outside cell}/{ion inside cell}
