Amixture of 15.0 g of the anesthetic halothane (c2hbrclf3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 torr. what are the partial pressures of each gas? a. phalothane = 778 torr, po2 = 84 torr b. phalothane = 162 torr, po2 = 700 torr c. phalothane = 84 torr, po2 = 778 torr d. phalothane = 155 torr, po2 = 707 torr e. phalothane = 707 torr, po2 = 155 torr.

Answer : The partial pressure of and are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of = 15.0 g

Mass of = 22.6 g

Molar mass of = 197.4 g/mole

Molar mass of = 32 g/mole

First we have to calculate the moles of and .

and,

Now we have to calculate the mole fraction of and .

and,

Now we have to partial pressure of and .

According to the Raoult's law,

where,

= partial pressure of gas

= total pressure of gas

= mole fraction of gas

and,

Therefore, the partial pressure of and are, 84 torr and 778 torr respectively.

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