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At 80 degrees celsius the vapor pressure of benzene is 753 torr and that of toluene is 290 torr. What is the total pressure and

nexus9112 [7]

Answer:

The total pressure of the solution is 598.66 torr

while the vapor pressure of the components is vapor pressure of benzene is =502 torr and that of toluene = 96.6 tor

Explanation:

To solve this question, we need to look at the known variables, the unknown variables, and then we select the appropriate relations between the known variables and the unknown

Here we have the temperature of the gases t = 80°C

the pressures of benzene = 753 torr

the pressures of toluene = 290 torr

mole fraction of benzene in solution = 2/3

mole fraction of toluene in solution = 1/3

the unknown variables = composition of the solution and the total pressure of the solution

From here given the known variables and the required variables, the appropriate relation bstween the known and unknown variables is Route's Law

Raoult's relates the vapor pressure of a mixture of gases to the mole fraction of solute gases introduced in the gas solution. Raoult's Law is expressed by the formula: Psolution = Χsolvent1×Psolvent1 + Χsolvent2×Psolvent2+ ... .

where. Psolution = vapor pressure of the solution and

Χsolvent1 = mole fraction of solvent 1

Psolvent1 = vapor pressure of solvent1

Thus we have by Raoult's Law

$P_{solution}$ = $X_{benzene}$ × $P_{benzene}$ + $X_{toluene}$ × $P_{toluene}$

which is =2/3×753 + 1/3×290 = 502 + 96.6 = 598.66 torr

and composition vapor pressure of benzene is 502 torr

while the composition vapor pressure of toluene is 96.6 torr

6 0

3 years ago

A pebble is dropped from rest from the top of a tall cliff and falls 53.4 m after 3.3 s has elapsed. How much farther does it dr

jeka94

Answer:

426.84 m

Explanation:

initial velocity u = 0

time t = 3.3 s

distance travelled s = 53.4 m

acceleration due to gravity = g

s = ut + 1/2 g t²

53.4 = 0 + 1/2 g x 3.3²

g = 9.8 m /s²

For the whole length of fall

distance travelled = h

total time = 6.6 + 3.3 = 9.9 s

h = ut + 1/2 g t²

u again = 0

h = .5 x 9.8 x 9.9²

= 480.24 m

distance travelled in last 6.6 s

= 480.24 - 53.4

= 426.84 m

6 0

2 years ago

List the ocean floor features that are formed by the movement of tectonic plates

Andreas93 [3]

Tsunami and under water volcano

8 0

3 years ago

a 3,000kg truck runs into the rear of a 1000kg car that was stationary. The truck and car are locked together after the collisio

joja [24]

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7 0

3 years ago

If two particles have equal kinetic energies, are their momenta necessarily equal? explain.

Mandarinka [93]

Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

$K.E=\frac{1}{2}mv^{2}$

Similarly the momentum is given by $m\times v$

For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective kinetic energies is given by

$K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}$

$K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}$

Similarly For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective momenta are given by

$p_{1}=m_{1}\times v_{1}$

$p_{2}=m_{2}\times v_{2}$

Now since it is given that the two kinetic energies are equal thus we have

$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)$

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0

3 years ago