The previous answers to the questions above were not correct and I would like th

Question

The previous answers to the questions above were not correct and I would like the correct answers with explanations to them as well as the question below

The U.S. Food and Drug Administration lists dichloromethane (CH2CI2) and carbon tetrachloride (CCl4) among the many cancer-causing chlorinated organic compounds. What are the partial pressures of these substances in the vapor above a mixture containing 1.45 mol of CH2Cl2 and 1.65 mol of CCl4 at 23.5° C? The vapor pressures of pure CH2Cl2 and CCI4 at 23.5° C are 352 torr and 118 torr, respectively. (Assume ideal behavior.) torr CH2 Cl2 CH2Cl2 =1214 Pcc46.22 torr CCl

Explanation / Answer

Given:

Moles of CH2Cl2 = 1.45

And moles of CCl4 = 1.65

Vapor pressure of pure CH2Cl2 = 352 torr

Vapor pressure of pure CCl4 = 118 torr

Now we know the vapor pressure of the substance = mole fraction x pure vapor pressure.

(Raults law)

Calculation of mole fraction of CH2Cl2

Mol fraction = mol CH2Cl2 / total moles

= 1.45 / (1.45+1.65 ) = 0.47

Mol fraction of CCl4 = 1.65 /(1.45 + 1.65)

= 0.53

Now vapor pressure of CH2Cl2 = 0.47 x 352 torr = 164.7 torr

Vapor pressure of CCl4 above the mixture = 0.53 x 118 = 63 torr

Q. 2

We are given KCl with 725 mass percent

Freezing point is = -0.340 0C

Normal freezing point of water (solvent) is 0 degree C so delta Tf = 0.340 deg C

Now we know the formula for depression in freezing point

Delta Tf = i x m x kf

Here the solution is of KCl and water.

m is molality.

So we need to calculate number of ions that are formed from KCl

Number of ions = 2 ( K+ and Cl-)

Now we use kf value of water and find out m

kf of water = 1.86 deg/ m

Lets plug this value in order to get molality.

m = Delta Tf / ( i kf )

= 0.340 deg C / ( 2* 1.86 deg C /m)

= 0.091398 m

So the molality of this solution would be = 0.0914 m

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