CONCENTRATION OF SOLUTIONS CHEMISTRY (MOLARITY)

Question

1. What is the molarity of a solution that contains 25 g of KOH in 250 mL of solution? 



1. What is the mole fraction of sulfuric acid (H2SO4) in 8% aqueous H2SO4 solution?

Answer 1

What is the molarity of a solution that contains 25 g of KOH in 250 mL of solution?

Use these formulas below. Find moles, find liters, then use those to find molarity.

Molarity=moles/Liters

M=n/v

Moles=mass/molarmass

n=g / (g/mol)

KOH, molar mass is K+O+H=39.10+16.00+1.01=56.11 g/mol

Calculate moles KOH:

n=25g / 56.11g/mol

n=0.4456 mol KOH

250mL of solution is 0.250 L solution.

250mL (1 L / 1000 mL)=0.250 L

M=n/v

n=0.4456 moll KOH

v=0.250 L

M=0.4456/0.250

=1.78 M KOH

1. What is the mole fraction of sulfuric acid (H2SO4) in 8% aqueous H2SO4 solution?

Use the percent by mass to get the grams of solute and solvent.

100g solution = 8 g H2SO4 + 92 g H₂O

Use molar masses to convert g to mol.

8 g H2SO4

molar mass H2SO4=98.072 g/mol

mol H2SO4 = 8/98.072

mol H2SO4 = 0.08157

92 g H₂O

molar mass H2O = 18.02 g/mol

mol H2O = 92/18.02

mol H2O = 5.105

write the mol fraction formula.

mol fraction of H2SO4 = ^{mol H2SO4}/_{total moles}

find total moles.

total moles = mol H2SO4 + mol H2O

total moles = 0.08157+5.105

total moles = 5.187

calculate mol fraction.

mol fraction of H2SO4 = ^{0.08157 mol H2SO4}/_{5.187 total moles}

mol fraction H2SO4 = 0.0157