Vapour pressure problem (simple)

In summary, to lower the vapour pressure of pure chloroform by 13.3 kPa at 70.0oC, 344 grams of iodine (I2) should be dissolved in 1 L of chloroform. This is calculated using Raoult's Law and the given information about the density and vapour pressure of chloroform.
  • #1
geffman1
67
0
The vapour pressure of pure chloroform at 70.0oC is 135.7 kPa. What mass of iodine (I2) should be dissolved in 1 L of chloroform, CHCl3 (density = 1.49 g/cm3) to lower the vapour pressure by 13.3 kPa?


hey guys i really can't do this question, i think in involves raults law, P=XPA
anyhelp would be good. thanks



p.s. answer is 1.32 x 101 g
 
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  • #2
No idea if you are to take into account iodine pressure or not.

Assuming it has to be ignored - at what x the pressure of chloroform vapour will be lowered by 13.3 kPa?
 
  • #3
thnks for the reply, umm 122.4kpa
 
  • #4
Hey,

Here is the solution I got, its different to your answer though...

Psolution=xsolvent.Psolvent

so Psolution=135.7-13.3 kpa (the new pressure of the solution)

now in 1L =1000cm3 there is 1490 grams = 12.482 moles

so re-arrange the equation and you get=> x = (12.482 - 11.26)/(3.55x10-3)

=344.23 grams

if 133.32 is the correct answer as you suggest I don't know where went wrong.
 
  • #5
hey mate i checked with the teacher and ur right(stupid internet tests). just wondering if you could show me where you got a few of the those numbers from, I've tried but get lost. thanks
 
  • #6
Ok,

so Xsolvent=molsolvent/(molsolute+molsolvent), we are looking to find molsolute (I2) so we can multiply by the RFM (253.8 g.mol-1) to get the mass needed.

1L of chloroform has mass 1490g = 12.482mol

Xsolvent=(135.7kPa-13.3kPa)/135.7kPa=.902

Xsolvent=12.482/(12.482+x)=.902

re-arranged x=1.356 mol

therefor mass I2 = 344g
 

Related to Vapour pressure problem (simple)

1. What is vapour pressure?

Vapour pressure is the pressure exerted by a gas when it is in equilibrium with its liquid or solid form. It is the measure of how easily a substance evaporates and becomes a gas.

2. How is vapour pressure calculated?

Vapour pressure can be calculated using the Clausius-Clapeyron equation, which takes into account the temperature and enthalpy of vaporization of a substance. It can also be determined experimentally using a device called a vapour pressure osmometer.

3. What factors affect vapour pressure?

The main factors that affect vapour pressure are temperature, intermolecular forces, and the surface area of the substance. As temperature increases, so does vapour pressure. Stronger intermolecular forces and a larger surface area lead to a higher vapour pressure.

4. How does vapour pressure affect boiling point?

Vapour pressure and boiling point have an inverse relationship. As vapour pressure increases, the boiling point of a substance decreases. This is because at higher vapour pressure, the molecules of the substance are more likely to escape from the liquid and become a gas, causing the substance to boil at a lower temperature.

5. Why is understanding vapour pressure important?

Understanding vapour pressure is important in various scientific fields, such as chemistry, physics, and meteorology. It helps in predicting the behavior of substances, such as their boiling points and phase changes, and is also important in industrial processes, such as distillation and evaporation. In addition, it is crucial in understanding the water cycle and atmospheric processes, such as cloud formation and precipitation.

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