If you need to make just 100mL, then you need 1 tenth of a liter that is 5M. If you were to make 1L of 5 molar NaCl, you would need 5 times the molar mass of NaCl (58.44g/mol) dissolved in 1L of water. Thus for 1L of a 5M solution you need 5 * 58.44g, or 292.2 grams of NaCl.
However, since we only want 100mL, which is 1/10 of a Liter, we also only need 1/10 the amount of NaCl, or 292.2 / 10, which is 29.22g.
So, measure out 29.22g NaCl, and dissolve completly in a volume less than 100mL, say 80mL, then bring the final volume up to 100mL.
You now have 100mL of a 5M NaCl solution.
MW of NaCl is 58.44
Dissolve 58.44g of NaCl in 1L of ddH2O will give 1 M of NaCl Solution
If u wish to prepare 500ml, then 29.22g of NaCl should be added...
250ml=.250L
.250L X .707M= .177 moles
.177 moles X 85g(molar mass)= 15.05g
Add 15.05 grams of NaNO3 to 260ml of solution in order to a .707 M solution
I would dissolve .4g of sodium thiosulfate into 250.0mL of water.
prepare of a 100ml of 0.2 NaOH
calculate the osmolarity of a solution ; 1m of sucrose at 25 degree centigra 2m kcl at 25degree centigre 154mM Nacl at 25 degree
1m x 1m x 1m = 1m3
Preparing 1N HCl for 1L. 1N=1M in HCl. Conc. HCl= 12M M1V1=M2V2 12*V1=1*1000 V1=1000/12 V1=83.33ml 1N HCl= 83.33ml of Conc. HCl in 1L of water 2N HCl= 167ml of Conc. HCl in 1L of water.
A 1M solution of sodium carbonate contains 1 gram formula mass of sodium carbonate dissolved in each liter of solution.
0.056g
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
A solution of NaCl 1 M.
C1V1 = C2V2 (500ml)(1M) = (Xml)(6M) ( divide both sides by 6 in this case ) = 83.3ml Put this 83.3ml of 6M acid into a 600ml beaker and add 416.7ml to get 500ml total.
Increase the concentration of NaCl evaporating the water.
1M contains 43.1ml in 500ml, 2M contains 86.2ml in 500ml
Let's see. NaOH + HCl = NaCl + H2O The usual salt ( NaCl ) and water.
10.82
Dilute 1M Tris 1:100
100ml / 1000 = 0.1mol *1M = divide by 1000 moles = mass / molecular mass mass = mole x molecular mass of nacl mass = 0.1 x 58.5 mass = 5.85 answer is 5.85 grams
Infinite because Sodium Chloride is neutral and will not neutralize sulfuric acid.
1m*1m*0.5m=0.5m3
40 grams, this is the 1M NaOH standard laboratory solution.