how many grams of salt (NaCl) will make 0.9% solution in 1 L of water?
913013
0.9% NaCl means 0.9 g NaCl in 100 mL SOLUTION, so you need 9 g NaCl for 1000 mL (1L)\n\nTake a volumetric cylinder, putsome water in it, dissolve the salt and add water to a final volume of 1L and mix. If you just add 9 g of NaCl in 1L of water the final volume of the solution will be different than 1L and thus your concentration will deviate from 0.9%.
694959
easy\nyou should have written (to a solution of 5% salt)\nthe quantity of salt in 6 grams of a 5% solution is .3 grams.\n\nthe amount to add to make a 30% solution can be calculated as x\n(,3 +x) /6= 30%\n.3 +x=1.8\nx= 1.5 grams of salt
1115475
To make 100mL: \n\n0.6% from a 10% solution is 6ml of the 10% diluted in 94ml of H2O.  \n\n0.6% from a 25% solution is 2.4ml of the 25% diluted in 97.6ml of H2O.\n\nThe math?\n\n0.6%*100mL / 10% = 6ml\n\n0.6%*100mL / 25% = 2.4ml
248635
To neutralize and  to have a pH of 7 you will need to add 200.g of NaOH \n\n Molar mass is 58.44g NaCl\n \nmass salt Concentration is 292.2g NaCl\n\nW% = 0.286%\n\nif, the M NaOH solu. is equal to  HCl solu.  of  .05M
603847
if its 2% by mass, thats 2 grams of KCl for every 100 grams of solution its in. \n\n2g KCl= 2g / (74.55g/mol)KCl = 0.0268mol KCl\n\nBecause there is one mole of Cl in KCl, this is also the number of moles of Cl in 100 grams of solution.\n\n0.0268mol Cl * (35.45g/mol)Cl = 0.951gCl/100gsolution\n\nThis is 0.951%\n\nnote: 35.45g/mol is the molar mass of Cl and 74.55g/mol is the molar mass of KCl. These were obtained from the periodic table.
237268
.001565mol / .075L = .021M NaCl
342141
ok well, the equation is balanced so you need equal moles of Ag and Cl to make the products. They give you the concentration and volume of Ag, so you can find the amount of moles by multiplying 0.104mol/L * 0.0272L = 0.0028288 mols of Ag. so 0.0028288 mols of Cl were there originally. and you have the volume of water, so you divide moles by volume and you get the concentration. 0.0028288 mols/0.025L = 0.113152 M. \n\nanother way to do it is the easy way, use c1v1 = c2v2. you have c1 = 0.104M, v1 = 27.2mL, v2 = 25mL, so you can find c2. you dont have to worry about units in this case so its easier. \n\nhope this helped. let me know if u need more help. laterz
862256
Going from 10^12 cells/mL to 10^10 is a 1:100 dilution.\nIf you want a final of 2mL, you need to add 20uL of your stock solution.\nSo do 20uL of stock (10^12)+ 1980uL of salt solution = 2000uL (10^10), which is the same as 2mL.
710835
Suppose that we have r milliliters of the 8% solution.  This means the amount of salt in it is .08r.\nSuppose we have s milliliters of the 3% solution.  This means that the amount of salt in it is .03s. \nWhen we mix them together we have r + s milliliters total, and the amount of salt in it is .08r + .03s.  If the resulting mixture is a 5% solution then this means that  \n(.08r + .03s)/(r + s) = .05.\nSo .08r + .03s = .05(r + s) = .05r + .05s.  \nThen .03r = .02s, or r/s = .02/.03 = 2/3. \nBut we know that s = 600 mL, so r = (2/3)s = (2/3)600 = 400 mL\n\nAdd 400 mL of the 8% solution to 600 mL of the 3% solution and get 1000 mL of 5% solution.
132728
I'm assuming that u r working with water as the solvent.\n\nDensity of water = 1 g/ml\ni.e. 1 kg of water occupies 1000 ml.\n\nAlso, 1 mg = 1000 ug\nSo, in fact, 1 mg/kg = 1 ug/ml, no need for any conversion factor.
776525
Since you are pretty, here's how:\n\n1. Compute for the number of moles NaOH needed to prepare the solution.\nmole NaOH: 1 L * 0.5 mol/L = 0.5 mol\n\n2. Get the weight NaOH you need by multiplying the number of moles to the molecular mass of NaOH.\nweight NaOH: 0.5 mol * 40g/mol = 20g\n\n3. To prepare, weigh 20 grams of NaOH pellets. Dilute this with water to 1000 mL in a volumetric flask.
775006
54.129 g -  42.889 g = 11,240 g of solution\n44,079 g - 42.889 g = 1,190 grams of dy salt\n%m/m= (1,190g salt/11,240g solution)*100= 10,59%\nThe density of 10%m/m solution of salt is aprox. 1,08g/mL\ntherefore\n11,240g solution*(1mL/1,08g)= 10,407 mL\n%m/v= (1,190g salt/ 10,407mL)*100= 11,43%\nmoles of salt=\n1,190g NaCl*(1mol NaCl/58,45g NaCl)= 2,036x10^-2 moles NaCl\nvolume:\n10,407 mL*(1liter/1000mL)= 1,0407x10^-2 L\nM= moles of salt/liters of solution\nM = 2,036x10^-2 moles NaCl/1,0407x10^-2 L= 1,956 M
202477
I'm not sure if I understand your question.\n\nIf you are wanting to make a 0.92 M NaCl solution out of 0.39 M NaCl, well you are going to have to work hard.  Basically, you would be boiling the water and having to measure the mass of the flask.  This is a very tedious process.  Bottom line, you cannot use the dilution formula to make it happen.\n\nNow, if you are wanting to make a 0.39 M NaCl solution from a 0.92 M NaCl solution, then you use:\n\nM1V1 = M2V2\n\n0.92 * V1 = 0.39 * (.1)\n\nremember .1 L is 100 ml.\n\nSolve for V1\n\nV1 = 0.42 L\nV1 = 42 mL\n\nSo, take 42 ml of the 0.92 M NaCl and dilute it up to 100 ml.\n\nHope it helps!
516306
Until the concentration of salt is large enough to form a saturated solution, ie. one in which no more solute will dissolve, the volume of the solvent (water) will not change.  20g of salt is not nearly enough to form a saturated solution, so the water level will remain at 500ml.
885670
Get the chemical equation first:\nHCl + NaOH --> NaCl + H2O\n\nMoles of HCl used = (50/1000)dm3 X (1.0mol/dm3) = 0.05 mol\nMoles of NaOH used = (50/1000)dm3 x (1.0mol/dm3) = 0.05 mol\n\nFrom chemical equation, 1 mole of HCl needs 1 mole of NaOH for complete neutralization.\nFrom what you are given, you have the exact amount for complete neutralization (meaning, there is no excess reagents).\nFrom chemical equation again, 1 mole of HCl or NaOH produces 1 mole of water.\nThus you had exactly 0.05 mole of water produced here.\nHope this helps. :)
64810
Assuming mol 1-L  = moles per liter \n\n8.9 x MW\n\nwhere MW = molecular weight of the compound
838309
b a solution, the salt will dissolve