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

Mole fraction H₂O → 0.72

Mole fraction C₂H₅OH → 0.28

Explanation:

By the mass of the two elements in the solution, we determine the moles of each:

25 g . 1 mol/ 18g = 1.39 moles of water (solute)

25 g . 1 mol / 46 g = 0.543 moles of ethanol (solvent)

Mole fraction solute = Moles of solute / Total moles

Mole fraction solvent = Moles of solvent / Total moles

Total moles = Moles of solute + Moles of solvent

1.39 moles of solute + 0.543 moles of solvent = 1.933 moles → Total moles

Mole fraction H₂O = 1.39 / 1.933 → 0.72

Mole fraction C₂H₅OH= 0.543 / 1.933 → 0.28

Remember that sum of mole fractions = 1

Mol fraction H2O = 0.72

Mol fraction C2H5OH = 0.28 moles

Explanation:

Step 1: Data given

Mass of H2O = 25.0 grams

Molar mass H2O = 18.02 g/mol

Mass of C2H5OH = 25.0 grams

Molar mass C2H5OH = 46.07 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles H2O = 25.0 grams / 18.02 g/mol

Moles H2O = 1.39 moles

Moles C2H5OH = 25.0 grams / 46.07 g/mol

Moles C2H5OH = 0.54 moles

Step 3: Calculate total moles

Total moles = 1.39 moles + 0.54 moles

Total moles = 1.93 moles

Step 4: Calculate mol fractions

Mol fraction = moles / total moles

Mol fraction H2O = 1.39 moles / 1.93 moles

Mol fraction H2O = 0.72

Mol fraction C2H5OH = 0.54 moles / 1.93 moles

Mol fraction C2H5OH = 0.28 moles

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol