What is the mole fraction of ethanol when 23.2 g of ethanol(C2H5OH) are mixed with 73.1 g of water(H2O)? (atomic mass of C= 12, H = 1, O = 16,)

0.111

Explanation:

Step 1: Calculate the moles corresponding to 23.2 g of ethanol

The molar mass of C₂H₅OH is 46 g/mol.

23.2 g × 1 mol/46 g = 0.504 mol

Step 2: Calculate the moles corresponding to 73.1 g of water

The molar mass of H₂O is 18 g/mol.

73.1 g × 1 mol/18 g = 4.06 mol

Step 3: Calculate the total number of moles

n = nC₂H₅OH + nH₂O = 0.504 mol + 4.06 mol = 4.56 mol

Step 4: Calculate the mole fraction of ethanol

We will use the following expression.

Χ(C₂H₅OH) = nC₂H₅OH / n

Χ(C₂H₅OH) = 0.504 mol / 4.56 mol = 0.111

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 :