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  • #16
    Originally posted by Alex View Post




    JohnS,

    I have a 80% Sodium Chlorite powder, what amount of it would I need to convert to 25% and how much water would I need to add to make a final 25% NaClO2 in 100 mL?

    What formula would one use to convert 80% to 25% and how to calculate the amount of water for a net 100 mL of 25%.

    Thanks.
    I assume you mean mass/volume percent as before so 250 g/L of NaClO2. You only want 100 mL, so you need 25 g NaClO2. However, it is only 80% pure, so you need to weight out 25 g/0.8 = 31.25 g of your powder. Dilute it to 100 mL final volume. As we have discussed before: Don't measure the water, measure the final volume if you want any precision.

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    • #17
      Originally posted by JohnS View Post

      I assume you mean mass/volume percent as before so 250 g/L of NaClO2. You only want 100 mL, so you need 25 g NaClO2. However, it is only 80% pure, so you need to weight out 25 g/0.8 = 31.25 g of your powder. Dilute it to 100 mL final volume. As we have discussed before: Don't measure the water, measure the final volume if you want any precision.

      Correct, stock 80% powder is in 1 kg/1000 grams glass bottle.

      Care to tell how did you come up with 250 g/L of NaClO2?

      Thanks.
      Last edited by Alex; 09-10-2020, 10:05 AM.

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      • #18
        Originally posted by Alex View Post


        Correct, stock 80% powder is in 1 kg/1000 grams glass bottle.

        Care to tell how did you come up with 250 g/L of NaClO2?

        Thanks.
        We have previously discussed how concentrations in percent are confusing and meaningless unless the basis is stated mass/mass, volume/volume, mass/volume, molar. I avoid the terms. Earlier we established you meant a mass/volume concentration, and in that case, percent is of a 1 kg mass per liter of solution so 25% means 250 g/L, which is clearer than stating 25%. Other possible meanings of 25% are 250 g/kg, 250 mL/L, and 250 mmol/mol. Those are generally preferred to saying 25% without specifying a basis.

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        • #19
          Originally posted by JohnS View Post

          We have previously discussed how concentrations in percent are confusing and meaningless unless the basis is stated mass/mass, volume/volume, mass/volume, molar. I avoid the terms. Earlier we established you meant a mass/volume concentration, and in that case, percent is of a 1 kg mass per liter of solution so 25% means 250 g/L, which is clearer than stating 25%. Other possible meanings of 25% are 250 g/kg, 250 mL/L, and 250 mmol/mol. Those are generally preferred to saying 25% without specifying a basis.
          Understood it's grams/grams.

          Thanks again.

          Comment


          • #20
            Originally posted by JohnS View Post

            We have previously discussed how concentrations in percent are confusing and meaningless unless the basis is stated mass/mass, volume/volume, mass/volume, molar. I avoid the terms. Earlier we established you meant a mass/volume concentration, and in that case, percent is of a 1 kg mass per liter of solution so 25% means 250 g/L, which is clearer than stating 25%. Other possible meanings of 25% are 250 g/kg, 250 mL/L, and 250 mmol/mol. Those are generally preferred to saying 25% without specifying a basis.

            Measured have 400 grams of 80% NaClO2 powder left.

            What amount of it would I require to make a 25% solution in 1000 mL/1 Liter?

            As per the formula 250 g/L.

            So it comes to 400 g/L= 400 g.

            400/ .8 = 500 g would be required to make a 25% solution in 1 liter.

            Since I don't have 500 g would need to reduce the amount?

            For example: 320 g /.8= 400 g.

            Accordingly would need to dissolve 320 vs 400 g in 1000 mL to give a net 25% solution.....is it correct?

            Thanks again

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            • #21
              As per another formula:

              400 g x 80% = 320 g weight of NaClO2.

              320 g/ 25% = 1280 g. Weight of 25% NaClO2 including water.

              1280 - 400 = 880 mL is the amount of water required to be added to make a 400 g 80% to 25%?

              Your expert views about it?

              Thanks again.

              It's confusing........
              Last edited by Alex; 09-11-2020, 10:25 AM.

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              • #22
                Do you know what you want? In post #29, I'm not sure if you want 1 L or all you can make from 400 g of impure powder. In post #21 you seem to have changed from w/v to w/w percentages. They are inherently different and can't be compounded or used interchangably.

                Post 20: Want 1 L. You need 250 g of NaClO2, but your powder is only 80% purity. You need 312.5 g of it to get 250 g to mix with sufficient water to make 1 L of mixture. If you want to use all 400 g of impure powder, you have 320 g of NaClO2 available, you can mix it to make 1.28 L of mixture, and 320 g/1.28 L = 25 g/dL or 25%. I n w/v compounding, you do not measure the water, you measure to the final volume of mixture.

                Post 21: Want 25% w/w. You have 320 g of NaClO2 available (in the form of 400 g of powder). At 25%, weight of total mixture is 1280 g. That is composed of 320 g of NaClO2. 80 g of impurities and 880 g of water. You can measure the weight of water added, or you can add water to achieve the total weight required. You use the solution by weighing it out and knowing 25% of the weight is NaClO2, you do not use it by measuring volume, because you don't know the volume. Weight/weight must be done on a scale. not in a measured volume.

                You have to sort out that there are multiple kinds of percent and it is a fuzzy term. It is better not used and replaced by expressing what you want with units. Do you want 25 g/dL or 25 g/hg, both of which are 25%, but different.

                Comment


                • #23
                  Originally posted by JohnS View Post
                  Do you know what you want? In post #29, I'm not sure if you want 1 L or all you can make from 400 g of impure powder. In post #21 you seem to have changed from w/v to w/w percentages. They are inherently different and can't be compounded or used interchangably.

                  Post 20: Want 1 L. You need 250 g of NaClO2, but your powder is only 80% purity. You need 312.5 g of it to get 250 g to mix with sufficient water to make 1 L of mixture. If you want to use all 400 g of impure powder, you have 320 g of NaClO2 available, you can mix it to make 1.28 L of mixture, and 320 g/1.28 L = 25 g/dL or 25%. I n w/v compounding, you do not measure the water, you measure to the final volume of mixture.

                  Post 21: Want 25% w/w. You have 320 g of NaClO2 available (in the form of 400 g of powder). At 25%, weight of total mixture is 1280 g. That is composed of 320 g of NaClO2. 80 g of impurities and 880 g of water. You can measure the weight of water added, or you can add water to achieve the total weight required. You use the solution by weighing it out and knowing 25% of the weight is NaClO2, you do not use it by measuring volume, because you don't know the volume. Weight/weight must be done on a scale. not in a measured volume.

                  You have to sort out that there are multiple kinds of percent and it is a fuzzy term. It is better not used and replaced by expressing what you want with units. Do you want 25 g/dL or 25 g/hg, both of which are 25%, but different.


                  Apologize for the confusion.

                  I measured and have 400 g of 80% purity NaClO2 powder. Want to bring it down to 25% in 1 Liter of water. What formula should I use?


                  What should be the quantity of 400g 80% NaClO2 powder and the amount of water required to be added to create a 25% NaClO2 solution in 1 liter.
                  Thanks.
                  Last edited by Alex; 09-12-2020, 09:21 AM.

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                  • #24
                    Originally posted by Alex View Post



                    Apologize for the confusion.

                    I measured and have 400 g of 80% purity NaClO2 powder. Want to bring it down to 25% in 1 Liter of water. What formula should I use?

                    Thanks.
                    Use 312.5 g of it and dilute in water to make 1 L final volume, as listed previously. 312.5 g x 80% = 250 g. Mixed to 1 L of mixture thats 250 g/L or 25% (w/v)

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                    • #25
                      Appreciate your expert response as always!

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