Question

A mixture of methanol and methyl acetate contains 15.0 weight percent methanol. Determine the number of gmols of methanol in 110.0 kilograms of the mixture.

Answer 1

There are 550.5 moles of methanol in 110.0 kilograms of the mixture.

Step-by-step explanation:

A solution 15% weight of methanol means there is 15g of methanol per 100g of the mixture or 0.1kg of the mixture. Also, the molar mass of methanol (CH3OH) is:

`m_(C) + 4xm_(H) + m_(O) = 12.0g/mol + 4x1.0g/mol +  16.0g/mol = 32.0g/mol`

Thus, dividing 15g by molar mass

`15.0g / 32.0(g)/(mol) = 0.5moles` we find there is 0.5 moles of methanol per 0.1Kg of the mixture. Calculating the number of mols of methanol in 110.0 kilograms of the mixture:

`(110.1Kgx0.5moles)/(0.1Kg) = 550.5 moles`

Therefore, there is 550.5 moles of methanol in 110.0 kilograms of the mixture.