Question

Ethanol, C2H5OH (molar mass = 46 g/mol) is mixed with methanol, CH3OH (molar mass = 32 g/mol) to make an ideal solution at a given temperature. If 5.00 g of ethanol and methanol are mixed, what is the resulting vapor pressure of the solution? (the vapor pressure at the same temperature for pure ethanol is 44.5 mm Hg and pure methanol is 88.7 mm Hg)

Answer 1

ANSWER:

Given,

mass of ethanol = 5 g

mola mass of ethanol = 46 g/mol

Number of moles of ethanol = {mass} / {molar mass} = {5 g} / {46 g/mol} = 0.109 mol

mass of methanol = 5 g

mola mass of methanol = 32 g/mol

Number of moles of methanol = {mass} / {molar mass} = {5 g} / {32 g/mol} = 0.156 mol

vapour pressure of pure ethanol, P^o_ethanol = 44.5 mmHg

vapour pressure of pure methanol, P^o_methanol = 88.7 mmHg

Now,

mole fraction of ethanol (_ethanol) = {no. of mole of ethanol} / {total number of mole}

= {0.109 mol}/{0.109 mol + 0.156 mol}

= 0.41

mole fraction of methanol (_methanol) = {no. of mole of methanol} / {total number of mole}

= {0.156 mol}/{0.109 mol + 0.156 mol}

= 0.59

Now,

According to Raoult's law,

P_ethanol = P^o_ethanol x _ethanol = 44.5 mmHg x 0.41 = 18.245 mmHg

P_methanol = P^o_methanol x _methanol = 88.7 mmHg x 0.59 = 52.333 mmHg

And,

Total vapour pressure of resulting solution, P = (P^o_ethanol x _ethanol) + (P^o_methanol x _methanol)

= P_ethanol + P_methanol

= 18.245 mmHg + 52.333 mmHg

= 70.578 mmHg

Hence, 70.578 mmHg is the resulting vapor pressure of the solution.