A medical company separates methanol (CH₃OH)) and ethanol (CH₃CH₂OH) by distillation. In one run, the starting solution was 25% methanol and 75% ethanol by mass at 25 $\circ$ C. In the liquid, the mole ratio of methanol to ethanol was ________, and in the vapor it was ________. The vapor pressures for methanol and ethanol at this temperature are 92 torr and 45 torr, respectively. A)0.48; 2.0 B)2.0; 0.48 C)0.33; 2.0 D)0.48; 0.33 E)2.3; 0.55

Question

Quizplus · Accepted Answer

The mole ratio in the liquid and vapor can be calculated using the given data. For the liquid phase, we first convert mass percentages to moles using the molar masses of methanol (32.04 g/mol) and ethanol (46.07 g/mol). For methanol: $ \frac{25\,g}{32.04\,g/mol} $ and for ethanol: $ \frac{75\,g}{46.07\,g/mol} $. The mole ratio of methanol to ethanol in the liquid is then calculated as the moles of methanol divided by the moles of ethanol, which gives approximately 0.48.For the vapor phase, we use Raoult's Law to find the partial pressures of methanol and ethanol, which are proportional to their mole fractions in the liquid phase and their vapor pressures. The mole fraction of methanol in the vapor phase is higher due to its higher vapor pressure compared to ethanol. The ratio of the partial pressures (which is also the ratio of mole fractions in the vapor phase) is given by $ \frac{92\,torr}{45\,torr} $, which simplifies to approximately 2.0. Therefore, the mole ratio of methanol to ethanol in the vapor is about 2.0.

A Medical Company Separates Methanol (CH3OH)) and Ethanol (CH3CH2OH) by Distillation

A Medical Company Separates Methanol (CH₃OH)) and Ethanol (CH₃CH₂OH) by Distillation