1.084g
use heat to heat the solution and add EDTA slowly to dissolve it.
You will be using the disodium salt of EDTA (M.W. = 372.24 g/mole). It has been dried for 1day at 80°C to drive off any superficial moisture. Transfer it in desiccator for an hour.Weigh carefully about 1.95 g of EDTA (record tothe nearest 0.1 mg). Quantitatively transfer this into a 500 mL volumetric flask then add 2-3 mLof pH 10 ammonia buffer. Fill the flask about halfway to the mark with deionized water andswirl to dissolve. This process can take up to 15 minutes. Once dissolved, dilute to the markand then cap and invert the flask at least 6 times to get a uniform solution. Keep the solutioncapped.This solution is about 0.01M but for an exact value you need Standardization of this solution with Calcium Carbonate standard solution.
About 80ml of water must be added to 40ml of a 25 percent by weight solution to make a 2 percent by weight solution.
If the percentages specified are by mass and the 900 ml of water are assumed to be at standard temperature and pressure, this problem can be solved as follows: The 900 ml of water will have a mass of 900 grams. Call the unknown mass of the 37 percent solution to be added m. this will contain 0.37m grams of formaldehyde and 0.63m grams of water, and 0.37m/(900 + m) = 0.10. Then, multiplying both sides by (m + 900), 0.37m = 0.10m + 90, or 0.27m = 90, or m = 3.3 X 102 grams of the 37 % solution to be added, to the justified number of significant digits.
You can't, at least not like that. 0.4 grams is an amountwhile percent is a fraction of a whole.To make 0.45 grams into percent you have to have something to compare it to, like 0.45 grams out of 0.90 grams is 50%.
use heat to heat the solution and add EDTA slowly to dissolve it.
To make a 9 percent saline solution, start by preparing a 100 percent salt solution. With a bottle of 100 percent salt water, take 9 percent and dilute with distilled water to make a 9 percent saline solution.
to make the reaction and associated calculations more complicated
It cannot be done since there is no substance such as NaCkl.
195 grams.
If your solution is a total of 414g and 3.06% of it needs to be NaCl, then you just take 414 x .0306 = grams of NaCl. The rest of the grams will be from other species in the solution.
Take 100 grams of 5% solution and do one of the following:Mix 95 g of it with 5 grams of sugar to end up with 100 g of 10% solution, or, when you are short of sugar:Evaporate 50 grams of water from 100 g of the 5% solution to end up with 50 g of 10% solution.
You dilute it 1:10, then you take 1 part of that solution and mix it with 9 parts of the diluent. That will make the 1:100 dilution you need, incl. prevention of pipette inaccuracy.
EDTA has a molecular weigh of 292.24g/mol. So if you want to make 1 litre for example of 0.5M EDTA, you'd weigh out 292.24*0.5 or 146.12g of EDTA and dissolve and fill to 1 litre with solvent
Put 100 grams in a beaker and and around 500 mls of water until it dissolves, then top up the beaker to a liter. That is your 10% solution. The percentage solution is a ratio of the weight of the compound to the weight of the final solution.
0.1M is 1/10 molar whereas 1mM is 1 millimolar and thus 1/1000 molar. There is thus a 1:100 dilution. So 10:1000 would be the same. To a 1000ml volumetric flask, pipete 10mls of 0.1M EDTA solution. Make up to the mark with deionized water. Mix and shake and you will have 1000mls of 1mM EDTA solution.
20% salt solution is the equivalent of adding 200gr salt in a 800 ml (1000ml -200ml) of water. you now have one liter of a 20% solution.