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At a given temperature the vapour pressures of pure liquid benzene and toluene are 745 torr and 290 torr respectively, A solution prepared by mixing benzene and toluene obeys Raoult’s Law. At this temperature the vapour pressure of benzene over a solution in which the mole fraction of benzene is equal to 0.340 is
a- 417 torr
b- 352 torr
c- 98.6 torr
d- 253 torr
e- none of these

Raoult’s law,
PA = PoA x XA
PB = PoB x XB

Dolton’s partial pressure law,
PT = PA + PB

PA = 745 X 0.34
Vapour pressure of benzene (PA) = 253.3 torr
The answer is (D)

PT = ( PoA x XA ) + ( PoB x XA )
PT = (745 x 0.34) + (290 x (1-0.34))
PT = 444.7 torr

At a given temperature the vapor pressure of pure liquid benzene and toluene are 745 torr and 290 torr, respectively. A solution prepared by mixing benzene and toluene obeys Raoult’s law. At this temperature the vapor pressure of benzene over a solution in which the mole fraction of benzene is equal to .340 is?

according to raoult’s law
Partial pressure due to a liquid in a mixture is equal to the product of its moles fraction in the liquid and its vapour pressure when it is pure.

Mole fraction of benzene = 0.34
Vapour pressure of pute benzene = 745 torr
Vapour pressure of benzene over the mixture = 0.34 x 745 = 253.3 torr

I don’t understand how to start this question. Is this Ideal gas law or am I way off?

First you need to know a couple of things:
Water gas is a mixture of CO and H2O.
Benzene is C6H6. It’s molecular weight is 78.11 g/mole.
So the reaction in question is:
2 C6H6 + 9 O2 = 12 CO + 6 H2O (g)
2 moles benzene produce 12 moles CO and 6 moles H2O, so
1 mole benzene produces 6 moles CO and 3 moles H2O, or a total of 9 moles of gaseous products.
9 moles of gas at STP = 9 (22.4 L) = 201.6 L
Those 9 moles of gas at 1.5 atm and 50 C = 201.6 x (1/1.5) x (323/273) = 159.0 L
So to produce 150 L at that temperature & pressure: 150/159 x 78.11 = 73.69 g benzene

Benzene diazonium chloride, C6H5NNCl, decomposes by a first-order rate law.

C6H5NNCl ? C6H5Cl + N2(g)
If the rate constant at 20°C is 4.3 10-5/s, how long will it take for 59% of the compound to decompose?

The first order rate law in integrated form is:
C/Co = e^-kt, where C is the amount of reagent REMAINING at any time t, and C=Co when t=0.
Then the equation becomes 0.41 = e^-(4.3×10-5 t)
Since e^-1 is about 0.39, then the exponent is slightly less, and t is about 2×10^4 sec.

Suppose that you have a solution of pure benzene. Use Raoult’s law to predict what would happen to the vapor pressure of the solvent, Psolvent, if you added some naphthalene (C_10_H_8_) solute.

a.) The vapor pressure of the solvent would decrease.
b.) The vapor pressure of the solvent would increase.
c.) The vapor pressure of the solvent would remain the same.
d.) You need more information to predict what would happen.

a.) The vapor pressure of the solvent would decrease.

The boiling point elevation of a solution consisting of 2.52 g of a nonvolatile solute in 89.7 g of benzene is 0.7928 oC. What is the molecular weight of the unknown solute?

0.7928 = kb x m = 2.53 x m

m = 0.313 = moles solute / 0.0897 Kg

moles solute =0.0281
MM = 2.52 g/ 0.0281 =89.8 g/mol

3. Which of the following pairs of molecules are expected to obey Raoult’s law?

a. CCl4, C6H6 (benzene)
b. CCl4, H2O
c. HF, H2O
d. CH3OH, H2O

the answer is a, c, and d..i m sure…jst got it rite…lol!

my father in law left the benzene in the garage and i smelled it for a couple of minutes.could that do any harm to the baby.i am 35 weeks pregnant.

I read about this on March of Dimes website and it said it is very dangerous but usually only if you are exposed to it often and for very long periods of time. Only smelling it for a couple of minutes probably didnt do anything, but if you want just call up your doctor to ask about it.

Benzene diazonium chloride, C6H5NNCl, decomposes by first order rate law. If the rate constant at 20 C is 4.3*10-5/s, how long will it take for 61% of the compaound to decompose? I saw someone already asked a similar question but the solution was hard to follow. Please help!

The integrated first order equation is:
ln[C6H5NNCl] = ln[C6H5NNCl]o – kt.
The left side of the equation is the concentration of the reactant at any time t, so if 61% of the C6H5NNCl decomposes, then 39% remains. The concentration on the right side of the equation is at time t = 0; so we can let [C6H5NNCl]o = 100 and [C6H5NNCl] = 39. Just substitute;
ln(39) = ln(100) – (4.3*10-5/s) t;
Solving, t = 21,899 s; or about 6 hr 5 min.

HN03 + Mg(OH)3—-> HOH + Mg(NO3)2

And could someone check these few please? Thanks

Benzene is a common hydrocarbon which is widely used as a solvent. Benzene is 92.3% carbon and 7.7% hydrogen.
a)What is the empirical formula for benzene? I got CH

b)From another experiment the molecular mass of benzene was determined to be 78 grams per mole. What is the molecular forumla for benzen? I came up with C6H6

Aspirin is 60% carbon, 4.48% hydrogen and 35.53% oxygen. What is the empirical forumla for aspirin? I came up with C9H8O4.

A) and b) are correct
Aspirin is correct

NOTE: it is Mg(OH)2 not Mg(OH)3
The balanced equation is-
2HNO3 + Mg(OH)2 –> 2H2O + Mg(NO3)2
Determine the molar mass of HNO3 (times 2) and Mg(OH)2 and add them together. Then determine the molar mass of H2O (times 2) and Mg(NO3)2 and add them together. Are the two numbers the same (the mass of the reactants and the mass of the products)?