What is the concentration in moles/litre that results from diluting a 10%w/v solution of NaCl?
965686
10% w/v could be 10g per 100ml of NaCl >> 10g/100ml\n1 mole of NaCl has 58.1g >> 1mol/58.1g\n\n\n[(10g NaCl)  (1 mole)] / [(100ml NaCl) (58.1 g NaCl)]\n\n1.72 M
732310
5.00/0.5 = 1 to 10 dilution\n\n600/10= 60 mL of stock needed\n\nthe rest (540mL) is water
939115
molarity (M)=(weight/mol.weight)*(1000/volume of the solution in mL)\n\nNow M = (27/17)*(1000//755)=2.1036 mol/lit
768198
molarity =moles/V where volume is expressed in L.\nmoles=mass/MW so\n\nmolarity=mass/(MW*V) =50/(106*0.5)=  0.94 M\n\nFor solutions, % (w/v) is grams of solute per 100 mL of solution.\nThus you have (50/500)*100% = 10%\n\n% is parts per hundred.\nso it is 10^4 times bigger than ppm\n so it is 10*10^4 ppm= 10^5 ppm.\n1 ppm corresponds to 1 mg of solute per 1L of solution.
1125846
NaVO3, MW = 122 g/mol\n\n100 mM = 0.1 M, so 0.1 moles in 1 L.  Dissolve 12.2g in water, dilute to 1L.
1104578
Use the equation M1V1 = M2V2\n\nM1 is the molarity of the HCL or 9.9 M\n\nV1 is the first volume of 102 ml but it needs to be changed to Liters. So it is 0.102L\n\nV2 is the total volume of the final solution so 525 ml (0.525 L) + V1 or 0.102L = 0.627\n\nM2 is what you are looking for - the final concentration or molarity.\n\nRearrange the equation to ((M1)(V1)) / (V2)  = M2\n\n(0.102L)(9.9M) / (0.627) = 1.61 M\n\nHope this helped!
676978
you can only dilute it w/v as it is a solid. Just take 10gr and dissolve it in about 50ml distilled water, then make up to 100ml when it has cooled.
681504
For this you need to the density or molarity of the solution.\n\nUsing molarity:\nMolarity=(Number of moles of solute)/(litre of solution)\n\nAlso 1 litre=1000 cubic centimeter.\n\nUsing density, d=m/V you get the mass of the solute.\nFrom there you can go to moles by using \nmoles=mass/(atomic or molecular mass)
784999
.15 M solution of NaCl means .15 moles of NaCl (solute) in 1000g of water (solvent).\ntherefore, to make 250 ml (or g) solution [ as density of H2O is 1]\nmoles of NaCl required = .15/4 = .0375 moles \n& moles = given wt * molecular wt \nso, given wt (or required wt in this case)= moles/ molecular wt. = .0375/58.1 = 645.4388 * 10^-6 g
16164
The instruction is not correct because the total volume yielded when you mix 100ml of concentrated solution and 900ml water will be somewhat less than 1000ml. The concentration will be slightly higher than 1.0M\n\nThe correct procedure is to take a 1l volumetric flask  and add the concentrated solution to about 800ml water, then fill up to the mark.
907422
Since Na+ and OH- ratio is 1 to 1 in the solution, the concentration of OH- would be 0.1M as well. \n\nC = n x V\n\nNumber of Na+ ions is equal to the number of OH- ions. So n is both same. If the volume is fixed, the concentration should be 0.1M.
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.
741946
Very easy:  \nPPM is equivalent to μg/mL (That is micrograms per milliliter)\n1 PPM = 1 μg/mL\n(the "μ" in micrograms is the greek letter mu)\n\nSo, a 10 PPM solution is 10 micrograms dissolved in 1 mL of water.  (or 10 milligrams dissolved in one liter (which is the same as10,000 micrograms in 1,000mL))\nSo, a 50 PPM solution is 50 micrograms dissolved in 1 mL of water.  (or 50 milligrams dissolved in one liter)\nEtc.\n\nPPM concentrations are just simple ratios, μg to mL.\n1 microgram = 0.000001grams\n\nThe volume (mL) always refers to WATER.  \nYou'll dissolve X amount of each of you reagents in X amount of water.
1048754
it depends on NaOH consentration\nit must be more than 7\nthis formula may help you\npH = 14 + log [NaOH]\n14 figure come from negative logatherm of Kw, 10^-14\n[NaOH]=NaOH concentration in mole/L
265173
You are correct using the formula:\n\nC1*V1 = C2 * V2\n\nwhere C is the concentration and V is the volume.\n\n\nNow, you know that you want a concentration of 0.39M solution of NaCl in a 100 ml sample.\n\nSo, the equation should read:\n\n0.82 * V1 = 0.39 * .1 L\n\n0.82*V1 = 0.039\n\nV1 = .047L\n\nso, take 47 ml of the .82M NaCl and dilute it up to 100 mL.  That will give you the 0.39 M NaCl.
83196
I've usually made up a concentrated stock of various buffers and dilute it to the concentration I need.  From a 1M Tris stock it's a 1/100 dilution to 10mM and from a 1M MgCLs it's a 1/1000 dilution.  I would decide how many milliters I wanted to make, measure out the cesium powder then pipette out the amount of each Tris (1/100 of the total solution vulume) and Mg (1/1000 of the volume) then bring to the final volume with distilled water.  This works fine if the concentrated solutions don't interact and cause precipitates or something else like being nearly a saturated solution.
959929
A.\nmole of HCl availible = 1.0 * 100 = 100 mmole\nmolarity of diluted HCl = 100 / 1000 = 0.10 M\n\nB.\nmole of HCl availible = 1.0 * 10 = 10 mmole\nmolarity of diluted HCl = 10 / 500 = 0.02 M\n\nC.\nmole of HCl availible = 1.0 * 50 = 50 mmole\nmolarity of diluted HCl = 50 / 100 = 0.50 M\n\npeace\nvixklen