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hieunguyen288 hieunguyen288
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10 years ago
What is the mole percent of methanol in your starting solution (composed of 1.5 ml methanol and 6ml isopropanol)
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#1
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10 years ago
Methanol density is 791,80 kg/m³ or 0,79180 g/ml. So, mass of methanol is 1.5*0,79180=1,1877g
Also, molar mass is 32.04 g mol−1, so M=m/n <=> n=m/M <=> n=1.1877/32.04 = 0,037 mol.  (rounded)

We do the same for isopropanol
Isopropanol density is 786,00 kg/m³ or 0,786 g/ml. So, its mass is: 6*0,786=4,716g.
Molar mass of isopropanol is 60,1 g/mol, so M=m/n <=> n= m/n <=> n=4,716/60,1 = 0,07846921797 = 0,078 mol (rounded)

So, the total mol are:0,037+ 0,078=0,115mol
And, the percentage of methanol mol is: 0,037 / 0,115=0,322 or 32,2% (rounded)



Please review the calculations.
hieunguyen288 Author
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#2
9 years ago
thank you
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