M V = M' V'
1M x 2L = 14.5 M x V'
V' = 138 mL
150 mL x 40.0 g LiNO3/100 mL solution = 60.g of solute
160 g (solution) - [5/100*160] g (solute) = 152 g (solvent) water
In this instance, 50 mol of sodium chloride is needed and molar mass of NaCl is 58.5 g/mol. Hence the mass we need is 29250 g. But this amount of salt could not be dissolved in 500 ml of water, so we cannot prepare this solution practically.
5mM = 0.005 moles 100 mL = 0.1 Liters Molarity = moles of solute/Liters of solution 0.005 M EDTA = X moles/0.1 Liters = 0.0005 moles EDTA =_____________ Now, look up the molecular formula for EDTA and find how many grams needed to add to your 100 mL.
(x L)((12%) = (100 L)(2%)x = 16.7 liters
1.17 grams :)
150 mL x 40.0 g LiNO3/100 mL solution = 60.g of solute
See the two Related Questions to the left for the answer.The first is how to prepare a solution starting with a solid substance (and dissolving it). The second question is how to prepare a solution by diluting another solution.
160 g (solution) - [5/100*160] g (solute) = 152 g (solvent) water
In this instance, 50 mol of sodium chloride is needed and molar mass of NaCl is 58.5 g/mol. Hence the mass we need is 29250 g. But this amount of salt could not be dissolved in 500 ml of water, so we cannot prepare this solution practically.
5mM = 0.005 moles 100 mL = 0.1 Liters Molarity = moles of solute/Liters of solution 0.005 M EDTA = X moles/0.1 Liters = 0.0005 moles EDTA =_____________ Now, look up the molecular formula for EDTA and find how many grams needed to add to your 100 mL.
100 M HCl don't exist.
To prepare 1 M CaI aqueous solution, dissolve 29.4 g in a total volume of 100 mls, or 294 g in a total volume of 1 liter.
(x L)((12%) = (100 L)(2%)x = 16.7 liters
1% solution = 1 gram per 100 mL, 10 grams per liter 20 grams
Dissolve 30g of sodium chloride in 100 mL of water.
1% 1g in 100ml= 1000mg in 100mlin 60ml ?100x = 1000 x 60100x = 60000x = 60000/100= 600150mg ?600/150 = 4answer = 4