Common commercial acetic acid is an aqueous solution that is 99% CH_3 CO_2 H by

Question

Common commercial acetic acid is an aqueous solution that is 99% CH_3 CO_2 H by mass, with a density of 1.05 g/cm^3. Calculate the molarity of this reagent. Molarity = mol/L Calculate the molality of this reagent. Molality = mol/kg Calculate the mole fraction of this reagent. Mole fraction = Common commercial ammonia is an aqueous solution that is 28% NH_3 by mass, with a density of 0.90 g/cm^3. Calculate the molarity of this reagent. Molarity = mol/L Calculate the molality of this reagent. Molality= mol/kg Calculate the mole fraction of this reagent. Mole fraction =

Explanation / Answer

Assume basis to be 1 kg of solution:

99 g of acetic acid and 1 g of water

In in this case , water is the solute and aa is the solvent

The molecular weight of AA =60.05 g/mol

Moles of AA=99 g/60.05 g/mol =1.648 mol

moles of water=1 g/ 18 g/mol=0.055 mol

Mole fraction of water = moles of water/Moles of water+moles of AA=(0.055 mol)/(1.648+0.055)mol=0.032

Mole fraction of AA=0.9677

Density of mixture =(density of AA*mole fraction of AA)+(density of water*mole fraction of water)

=1.05 g/cm^3*0.032+(1g/cm^3*0.9677)=1.0003 g/cm^3

Density of solution=1.0003 g/cm^3

mass of solution/Density of solution=volume of solution=100 g/1.0003 g/cm^3=99.87 cm^3 =0.0998 l

MOLARITY=moles of water/volume of solution in L=0.032 mol/0.0998 L=0.32 M

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