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

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the answer for sodium is 0.186

the answer for rubidium is 0.251