Asolution of phosphoric acid was made by dissolving 10.0 g h3po4 in 100.0 ml water. the resulting volume was 104 ml. calculate the density, mole fraction, molarity, and molality of the solution. assume water has a density of 1.00 g/cm3 .

density is 1.057 g/mL

mole fraction is 0.0180

molarity is 0.98 mol/L

mole fraction is 0.0180

Explanation:

Given data

acid mass = 10 g

water volume = 100 mL

total volume = 104 mL

density = 1 g/cm³

to find out

the density, mole fraction, molarity, and molality

solution

first we calculate the density that is = total mass g / volume of solution mL

total mass = mass of H3PO4 + water  mass

so water mass = density x volume

water mass = 100 ml x 1.0  g/cm3

water mass = 100 g  

so total mass  = 110.00 g

so that

density = total mass g / volume of solution mL

density = 110 / 104 = 1.057 g/mL

now we calculate no of moles in solvent i.e = mass  H2O / mlar mass H2O

no of moles in solvent = 100/18.015 = 5.55 moles

and no of moles in solute i.e = mass of H3PO4 / mlar mass in H3PO4

moles in solute i.e = 10/ 97.994 = 0.102 moles

so total moles is  5.55  + 0.102 = 5.652 moes

so now mole fraction = no of moles in solute / total moes

mole fraction = 5.55 / 5.65

mole fraction is 0.0180

now we calculate first

mole fraction in solute and solvent that is

mole fraction in solute = no of moles in solute/ total moles

mole fraction in solute = 0.102 /5.65

mole fraction in solute is 0.0180

and mole fraction in solvent that = no of moles in solvent/ total moles

mole fraction in solvent that = 5.55/ 5.65

mole fraction in solvent that is 0.982

so molarity = no of moles of solute / volume

molarity = 0.102/0.104

molarity is 0.98 mol/L

and molality  is = no of moles of solute / mass

molality = 0.102 / 100

molality  is  1.02 mol/kg