Question

A solution is prepared by mixing 25.0 g of water, H2O, and 25.0 g of ethanol, C2H5OH. Determine the mole fractions of each substance

Answer 1

The molar fraction of the "$H_2O \ and \ C_2H_5OH$" is "0.72 and 0.28"

Explanation:

We determine each mole by the mass of the two elements in the solution:

Calculating the moles of water (solute) $= \frac{1\ mol}{ 18 \ g} \times 25\ g = 1.39 \ moles$

Calculating the moles of ethanol (solvent ) $= \frac{1\ mol}{ 46 \ g} \times 25\ g = 0.543 \ moles$

$\text{Solute mole fraction} = \frac{Solute\ mole}{ Total \ moles}\\\\\text{Solvent mole fraction} = \frac{Solvent \ mole}{ Total \ moles}\\\\\text{Total moles = solute mole+ solvent moles}$

$1.39 \ solute \ moles + 0.543 \ solvent \ moles \ = 1.933 \ moles \to Total \ \ moles\\\\ \to H_2O \ Mole\ fraction= \frac{1.39}{ 1.933}=0.72\\\\ \to C_2H_5OH \ Mole \ fraction = \frac{0.543 }{1.933 }= 0.28$

Recollect that mole fractions total amount = 1

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