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How to calculate molality with density and molarity?

Molarity and molality help?

  • a) A solution of phosphoric acid was made by dissolving 10.5 g of H3PO4 in 100.0mL of water. The resulting volume was 104mL. Calculate the molarity of this solution. b) Assuming the density of water is 1.00g/mL, calculate the molality of the solution. c) What volume of a 0.67M sodium nitrate solution contains 1.44g of solute? I've got some answers, but I really would like some reassurance I got this right. Thank you.

  • Answer:

    The molar mass of H3PO4 is 98.0 g/mol so 10.5 g 0.1071mol; this is in 0.104 L of water, so the molarity is 0.1071/.104 = 1.03 M The molar mass of NaNO3 is 85.0 g/mol 1.44g = 1.44/85.0 = 0.0169 mol moles = conc*volume, so volume = moles/conc = 0.0169/0.67 = 0.0253 L = 25.3 mL

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a. mol H3PO4 = gr/Mr = 10,5 g / 98 g/mol = 0,107 mol M (molarity) = mol / L (volume of the solution on liter) = 0,107 mol / 0,104 L = 1,03 M b. It is relates with the question part a or not?? Sorry, I still study english but I know to do the questions like this ^,^ c. mol = gr/Mr = 1,44 / 85 = 0,017 mol M = mol / V 0,67 M = 0,017 mol / V V = 0,017 / 0,67 = 0,025 L So, the volume of a 0,67 M sodium nitrate solution is 0,025 L

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