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Calculate the molarity and molality of 37.0%(w/w) HCI with a density of
1.19 g mL-1.The molecular mass of HCI is 36.46 g mol-1.

  • chemistry - ,

    37.0% =37.0g/100g of mixture. 37g/36.46g mol-1 will give you the number of moles in your solution. Since 63% =63g of the remaining mixture, use the density and solve for the volume (63g/1.19g mL-1). Take the answer that you get for moles and the answer you get for volume and calculate the molarity (moles/volume). To calculate the molality, take the answer that you get for moles and divide it by the weight of solvent (moles/63g). I am a little rusty at performing chemistry calculations, so hopefully someone else comes along and verifies this or gives you the correct way of solving the problem. But I believe this is correct.

  • chemistry - ,

    M = mols/L, then
    1.19 g/mL x 1000 mL x 0.37 x (1 mol/molar mass HCl) = M = approximately 12M

    m = mols/kg solvent.
    How many grams of the solution do you have above? You know that's
    1.19 x 1000 = g soln.

    How many grams HCl are there?
    That's g soln x 0.37

    How many grams water (solvent) are there?
    That's g solution - g HCl = g solvent.
    Change to kg solent and
    m = mols HCl/kg solvent. I think it's something like 15 or 16 m.

  • chemistry - ,

    I forgot to add to the explanation that when calculating everything, all units for volume should be converted from mL to L, and the units for mass, when calculating molality, should be converted from g to Kg.

  • chemistry - ,

    Also, grams of the solution shouldn't be 63g but 100g.

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