if there is 1 milligram of a solute present in 1 liter of solution the concentration of the solution is ?
397236
1mg /L is called  "ppm" ---Parts per million\n\nie.,  1ppm solution having 1mg of solute in 1L solvent
1068727
Normality:\nnormality of a solution is defined as the number of gram equivalents of the solute present in one litre of the solution. it is denoted by "N".\n\nNormal solution:\na solution having normality equal to one is called a normal solution.\nSUCH  A  SOLUTION  CONTAINS  ONE  GRAM EQUIVALENT  OF  THE  SOLUTE  PER  LITRE OF  THE SOLUTION.\n\nMolarity:\nmolarity of the solution is defined as the number of gram mole of the solute present in one litre of the solution. it is represented by "M".\n\nmolar solution:\na solution having molarity equal to one is called molar solution.\nSUCH  A  SOLUTION  CONTAINS  ONE  MOLE  OF  SOLUTE  PER  LITRE  OF  THE  SOLUTION.
809824
If the concentration of the reactants is equal, it will be A. The smallest
39876
To answer this question you must look at your given data. 13% of 1.1 = .143 so you have .143g NaCl for every ml solution. Molarity is moles solute/liters solution. You just calculate how many grams NaCl are in a liter and convert to moles, do the division and your done.
168817
3L because 1L equals to 1000ml and 3000/1000 equals to my answer
684681
C because the solution is SATURATED with solute, meaning no more solute can be dissolved by the solvent in the solution.
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.
808705
Molarity of the solution = Number of moles of citric acid /1000 ml of solution\n\n= (250/192.1) moles of citric acid per 1000 ml of solution\n\n= 1.3014 M
74467
The solution has a density of 1.0 g per millilter, and it is a 0.1% concentration of Thimerosal.  So, the thimerosal has a concentration of 0.1% x 1.0g/mL, which is 1 mg/mL.\n\na) 0.5g of solution / 1.0 g/mL = 0.5 mL of solution\n0.5mL x 1.0mg/mL = 0.5 mg of thimerosal\n\nb) 50g of solution / 1.0 g/mL = 50 mL of solution\n50mL x 1.0mg/mL = 50 mg of thimerosal\n\nc) 25g of water / 1.0 g/mL = 25 mL of solution\n25mL x 1.0mg/mL = 25 mg of thimerosal\n\nd) 75mL x 1.0mg/mL = 75 mg of thimerosal\n\ne) 0.75g / 1.0mg/mL = 750 mg of solution\n750 mg / 1.0 mg/mL = 750 mL of solution
770958
1/1000 x molar mass (grams/litre) and solute in 1 litre
327703
The amount is diluted by a factor of 25.  So the resulting solution would have a concentration of 11.6M divided by 25 or \nor 0.464M.
984571
1 litre (1000 ml) of 1 M solution of NaCL contains 23 + 35.5 gms =58.5 gms (ie 1 molecular wt of sodium chloride in gms dissolved in 1 lit of solution)\n\nHence 1 lit of 0.360 M solution  will contain 0.360 X 58.5 gms or 21.06 NaCl (Sodium Chloride) \n\nTo find out exact vole in ml which contain 21.14 find volume by this formula 2.14 divide by 21.06 multiply 1000 , (ie 2.14/21.06 * 1000 ) 101.6 ml\n\nPS\nbasic assumption is 0.360 M indicates .360 Molar solution. \n(ie: M stands for Molar)
820326
If this is in Normality;\n\n0.1 Normal solution of  NaOH\n2 grams of NaOH  in a liter = 0.05 N\n\nN = weight in grams divided by MW (molecular weight ) = 40 / valence =1 / 1 liter. = 0.05\n\nMW = (Sodium) 23 + 16 (Oxygen) + 1 (hydrogen) = 40\n\nTherefore: Using the formula N1V1 = N2V2\n(0.1)( X) = (0.05) ( 1 liter)\n\nBy cross multiplication:\n\nX = (.0.05) ( 1) divided by ( 0.1) \nX =  0.5 liters or 500 milliliters\n\nbecause  1,000 ml = 1 liter\n\nYou will get the same answer for Molarity because the valence is one.\n\nso the answer is 500 milliliters.
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!
896601
Iwould estimate 1 milligram.
890187
There are 10 dL in one liter.\nif you have 0.42 mg in one liter, then you will have less mg in a smaller amount of solution, namely one deciliter.  Since one dL is 10 times smaller than one liter, there will be 10 times less mg in one dL. \nSo the answer is 0.042 mg
564420
.1ml of 100:1 dilution had 280\n1.0 ml of dilution has 2800\n1.0 ml of undiluted has 280,000