ure to answer all parts. The vapor pressure of ethanol (C2H5OH) at 20°C is 44 mmHg, and the vapor pressure of methanol (CH3OH) at the same temperature is 94 mmHg. A mixture of 25.3 g of methanol and 47.1 g of ethanol is prepared and can be assumed to behave as an ideal solution. Calculate the vapor pressure of methanol and ethanol above this solution at 20°C. Be sure to report your answers to the correct number of significant figures.

Explanation:

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In this case, by considering the Raoult's law we can write:

Whereas the purpose is to compute the pressures of both methanol and ethanol (pressures above the solution), thus, the first step is to compute the molar fractions:

Therefore, the pressures turns out:

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answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :