Question
At 21.0 °C, a solution of 23.71 g of a nonvolatile, nonpolar compound in 30.64 g of ethylbromide, C2H5Br, had a vapor pressure of 319.7 torr. The vapor pressure of pure ethyl bromide at this temperature is 400.0 torr. Assuming an ideal solution, what is the molecular mass of the compound?
Answers
pEtBr = XEtBr*P<sup>o</sup><sub>EtBr</sub>
319.7 = X*400
XEtBr = about 0.8 (estimated) see last post.
Xunk = 1-about 0.8 = about 0.2
mols unk = 23.71/molar mass or 23.71/Y.
mols EtBr = 30.64/109 = about 0.281
Then Xunk = mols unk/total mols.
0.2 = [(23.71g/Y)/(23.71Y)+(0.281)]
and solve for Y = molar mass unknown.
319.7 = X*400
XEtBr = about 0.8 (estimated) see last post.
Xunk = 1-about 0.8 = about 0.2
mols unk = 23.71/molar mass or 23.71/Y.
mols EtBr = 30.64/109 = about 0.281
Then Xunk = mols unk/total mols.
0.2 = [(23.71g/Y)/(23.71Y)+(0.281)]
and solve for Y = molar mass unknown.
How do I solve for the y? I get
0.2=23.71g / 24.71+.281
Y. Y
But I don't know how to solve
0.2=23.71g / 24.71+.281
Y. Y
But I don't know how to solve
Then Xunk = mols unk/total mols.
0.2 = [(23.71g/Y)/(23.71Y)+(0.281)]
and solve for Y = molar mass unknown.
<b>I made a typo in the above. It should be
0.2 = [23.71/Y/(23.71/Y)+ (0.281)]</b>.
First the numerator = 0.2*denominator.
23.71/Y = [0.2*(23.71/Y) + (0.281)]
23.71/Y = (4.74/Y)+ 0.0562
Multiply through by Y
23.71 = 4.74 + 0.0562Y
23.71-4.74 = 0.0562Y = 18.97
Y = 18.97/0.0562 = 337.5 = molar mass of unknown.
You can check that, and should.
molar mass ethyl bromide = about 109
molar mass unknownh = about 337.5
mols ethyl bromide 30.64/109 = 0.281
mols unknown = 23.71/337.5 = 0.0702
total mols = sum = 0.351
Then Xethylbromide = 0.281/0.351 = 0.800
and
Xunknown = 0.0702/0.351 = 0.200
so it checks.
0.2 = [(23.71g/Y)/(23.71Y)+(0.281)]
and solve for Y = molar mass unknown.
<b>I made a typo in the above. It should be
0.2 = [23.71/Y/(23.71/Y)+ (0.281)]</b>.
First the numerator = 0.2*denominator.
23.71/Y = [0.2*(23.71/Y) + (0.281)]
23.71/Y = (4.74/Y)+ 0.0562
Multiply through by Y
23.71 = 4.74 + 0.0562Y
23.71-4.74 = 0.0562Y = 18.97
Y = 18.97/0.0562 = 337.5 = molar mass of unknown.
You can check that, and should.
molar mass ethyl bromide = about 109
molar mass unknownh = about 337.5
mols ethyl bromide 30.64/109 = 0.281
mols unknown = 23.71/337.5 = 0.0702
total mols = sum = 0.351
Then Xethylbromide = 0.281/0.351 = 0.800
and
Xunknown = 0.0702/0.351 = 0.200
so it checks.
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