How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?
Molar mass ethanol:
C2H5OH = 12 x 2+ 1 x 5 + 16 + 1 = 46.0 g/mol
volume = 545 mL in liters: 545 / 1000 => 0.545 L
number of moles:
29.0 / 46.0 => 0.6304 moles
M = n / V
M = 0.6304 / 0.545
M = 1.156 mol/L
hope this helps!
Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
