Originally posted by Alex
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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; 09102020, 10:05 AM.
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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.
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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.
Thanks again.
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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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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; 09112020, 10:25 AM.
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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.
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Originally posted by JohnS View PostDo 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; 09122020, 09:21 AM.
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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.
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