I was wondering if I was doing a conversion right. I am supposed to take aqueous

Question

I was wondering if I was doing a conversion right. I am supposed to take aqueous HNO3 with a density of 1.42 g/mL and a mass percent of solution of 70% and convert that into molarity. Would the correct conversion be to take 70 g of HNO3 over 100 g of H2O multiplied by 1 mol HNO3 over 63.01 g HNO3 (the molar mass) and then multiply that by 1 g H2O over 0.001 L of H2O to get the moles per liter or would I use the given density (1.42 g/mL) from the problem? Also, would the molality be 11.1 mol HNO3/kg H2O and would the mole fraction be 1/6 mol HNO3/mol H2O

Explanation / Answer

Molarity = mass % x density x 10 / Molar mass

             = 70 x 1.42 x 10 / 63.01

              = 15.78 M

moles of solute = 70 g / 63.01 = 1.11 moles

mass of water = 100 g = 0.1 kg

molality = moles of solute / mass of solvent in kg

             = 1.11 / 0.1

molality = 11.1 mol HNO3 / kg H2O

moles of water = 100 / 18 = 5.56

mole fraction HNO3 = 1.11 / (1.11 + 5.56)

                                = 0.166

                                = 1/6 mol HNO3/mol H2O

Note : yes all are correct

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