The U. S. Food and Drug Administration lists dichloromethane (CH2Cl2) 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 1.80 mol of CH2Cl2 and 1.75 mol of CCl4 at 23.5°C? The vapor pressures of pure CH2Cl2 and CCl4 at 23.5°C are 352 torr and 118 torr, respectively. (Assume ideal behavior.)

P(CH₂Cl₂) = 178.5torr

P (CCl₄) = 58.2 torr

Explanation:

Assuming ideal behavior, the partial pressure is equal to mole fraction times vapor pressure of the pure substance. Partial pressure is:

Partial pressure CH₂Cl₂= X(CH₂Cl₂) * P°(CH₂Cl₂)

Partial pressure CCl₄ = X(CCl₄) * P°(CCl₄)

Mole fraction of both compounds is:

X(CH₂Cl₂) = 1.80mol / (1.80mol + 1.75mol) = 0.507

X(CCl₄) = 1 - 0.507 = 0.493

Solving, pressure of the solution:

P(CH₂Cl₂) = 0.507 * 352torr = 178.5 torr

P (CCl₄) = 0.493 * 118torr = 58.2 torr

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i