Question

Find the mole fraction of methanol and water in a solution prepared by dissolving 4.5g of alcohol in 40g of water​

Answer 1

Answer:

Explanation:

mass of alcohol (CH3OH) = 4.5g

molar mass of CH3OH = 32g/mol

so, mole of CH3OH, n = 4.5/32 mol

= 9/64 mol = 0.14 mol

again, mass of water = 40g

molar mass of water = 18g/mol

so, mole of water = 40/18

= 20/9 ≈ 2.22 mol

now mole fraction of methanol= mole of methanol/(mole of methanol + mole of water)

= 0.14/(0.14 + 2.22)

= 0.14/2.36

= 0.059 ≈ 0.06

hence, mole fraction of methanol is 0.06

Answer 2

Answer:-

• Mole fraction of methanol = 0.06

• Mole fraction of water = 0.94

Explanation:-

Here, methanol is solute and water is solvent.

• Molar mass of Hydrogen (H) = 1g/mol

• Molar mass of Oxygen (O) = 16g/mol

• Molar mass of Carbon (C) = 12g/mol

Hence, molar mass of methanol [CH₃OH] :-

= 12+1×3+16+1

= 12+3+17

= 32g/mol

And, molar mass of water [H₂O] :-

= 1×2+16

= 18g/mol

Number of mole in 4.5g of methanol :-

= Given Mass/Molar mass

= 4.5/32

= 0.140625 mole

Number of mole in 40g of water :-

= Given Mass/Molar mass

= 2.22 mole

Mole fraction of solute [methanol] :-

= Mole of solute/Total mole in solution

= 0.140625/[0.140625 + 2.22]

= 0.140625/2.360625

= 0.06

Mole fraction of solvent [water] :-

= Mole of solvent/Total mole in solution

= 2.22/[0.140625 + 2.22]

= 2.22/2.360625

= 0.94