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Analytical chemistry calculation mol/L to grams

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  • Analytical chemistry calculation mol/L to grams

    Hello everyone,

    I am stuck in a difficult situation. I do not know that how many grams will require to prepare 5.2*10^-2 mol/L AlNH4(SO4)2.12H2O‎ aqueous solution?

    Thanks.
    Last edited by sidj; 04-13-2020, 12:57 AM.

  • #2
    Originally posted by sidj View Post
    Hello everyone,

    I am stuck in a difficult situation. I do not know that how many grams will require to prepare 5.2*10^-2 mol/L AlNH4(SO4)2.12H2O‎ aqueous solution?

    Thanks.
    You have to look up, or calculate from the chemical formula, the molar mass. Wikipedia says it is 453.33 g/mol for the hydrate form. Then multiply by the molar concentration to get the mass concentration of 23.573 g/L. Then multiply by the volume you wish to make up.

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    • #3
      Thank you so much for your help.

      So it means

      if I have to prepare 100 ml of 5.2*10^-2 mol/L solution of AlNH4(SO4)2.12H2O,

      so MMAlNH4(SO4)2.12H2O = 453.32 g/mol

      g AlNH4(SO4)2.12H2O = 453.32 * 0.052 * 0.1

      AlNH4(SO4)2.12H2O =2.36 g


      I have one more question
      how to prepare 100 ml of 10 wt.% AlNH4(SO4)2.12H2O solution?

      Thank you.
      Last edited by sidj; 04-19-2020, 11:19 PM.

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      • #4
        To prepare 100 mL exactly, you need to know the density of the solution. You will prepare a few extra mL, but I recommend the following. Start with 100 mL of water, either weigh it or determine the density from temperature (slightly less than 1 g/mL). Measure out 1/9 of that weight in solute, and mix it. As 9 parts water, 1 part solute by weight, it will be 10%, but the volume will be a bit over 100 mL.

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