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How many moles of NaCl is required to make 1L of a 140mM solution?
To find the moles of NaCl needed, first convert 140 mM to M by dividing by 1000. Then, use the formula moles = Molarity x Volume (in liters) to calculate the moles required. In this case, the calculation would be: 0.140 mol/L x 1 L = 0.140 moles of NaCl.
How many moles of NaCl are contained in 100o ml of a 0.20 M solution?
Molarity = moles of solute/Liters of solutionMoles of solute = Liters of solution * Molarity ( 100 mL = 0.1 Liters )Moles of NaCl = 0.1 Liters * 0.20 M NaCl= 0.02 moles NaCl============
How many moles of sodium chloride are in 5.08 L of a 2.36 M solution?
To find the number of moles of sodium chloride, you can multiply the volume of the solution by its molarity. moles = volume (L) * molarity moles = 5.08 L * 2.36 mol/L moles = 11.9928 mol Therefore, there are approximately 11.99 moles of sodium chloride in 5.08 L of a 2.36 M solution.
What is the mole concentration of 58.44g of NaCl in 1 L of water?
Molarity = moles of solute/Liters of solutionGet moles NaCl.58.44 grams NaCl (1 mole NaCl/58.44 grams)= 1 mole NaCl------------------Molarity = 1 mole NaCl/1 liter= 1 M NaCl========
What is the molarity of a 2.000 L solution thatis made from 14.60 g of NaCL?
Find moles NaCl first.14.60 grams NaCl (1 mole NaCl/58.44 grams)= 0.2498 moles NaCl================Now,Molarity = moles of solute/Liters of solutionMolarity = 0.2498 moles NaCl/2.000 Liters= 0.1249 M NaCl solution--------------------------------
100ml of 0.200 molL NaCl is mixed with 400ml of 0.200 molL NaCl What is the concentration of Cl in the resulting solution?
The total moles of NaCl in the solution is 0.2 mol/L * (0.1 L + 0.4 L) = 0.3 moles. Since NaCl dissociates into 1 Na+ ion and 1 Cl- ion, the total moles of Cl- ions is also 0.3 moles. The concentration of Cl- ions in the resulting solution is 0.3 moles / (0.1 L + 0.4 L) = 0.6 mol/L.
What is the molarity of a 3000 liters solution containing 300 grams of NaCl?
Molarity = moles of solute/Liters of solution Find moles NaCl 300 grams NaCl (1 mole NaCl/58.44 grams) = 5.13347 moles NaCl Molarity = 5.13347 moles NaCl/3000 Liters = 1.71 X 10^-3 M sodium chloride ----------------------------------------
How do you find the total mass of NaCl in a 455000L pool at a constant concentration of 0.048 M?
You start by checking how many moles of NaCl you have in your pool. Number of moles=concentration*volume =0.048mole/L*455000L=21840moles Then you get the molar mass of NaCl which is 58.443 g/moles, then you multiply it by your number of moles. 21840 moles*58.443g=1'276'395g=1'276.395kg of NaCl
How many moles per mL of the solute are contained in the 0.20 M NaCl?
0.0002 mol NaCl/mLmolarity = (moles solute)/(L of solution)0.20 M NaCl = 0.20 mol NaCl/L1 L = 1000 mL0.20 mole NaCl/1000 mL = 0.00020 mol NaCl/mL (rounded to two significant figures)
What is the concentration M of a NaCl solution prepared by dissolving 3.8 g of NaCl in sufficient water to give 245 mL of solution?
Get moles NaCl and change 245 ml to 0.245 Liters. 3.8 grams NaCl (1 mole NaCl/58.54 grams) = 0.0650 moles NaCl Molarity = moles of solute/Liters of solution Molarity = 0.0650 moles NaCl/0.245 Liters = 0.27 M NaCl ----------------------
How many grams of NaCl are in 1.92 moles of NaCl?
Every mol of NaCl contains a mol of Na, weighting 23 grams, and a mol of Cl, weighing 35.5 grams. So, every mol of NaCl weights 58,5 (=23+35.5) grams. Therefore, 145 moles of NaCl weights 8482.5 grams.
How many mL of .117M AgNO3 solution would be required to react exactly with 3.82 moles of NaCl?
Balanced equation first! AgNO3 + NaCl -> AgCl + NaNO3 all one to one, get moles AgNO3 3.82 moles NaCl (1 mole AgNO3/1 mole NaCl) = 3.82 moles AgNO3 ------------------------------- Molarity = moles of solute/Liters of solution 0.117 M AgNO3 = 3.82 moles AgNO3/Liters Liters = 3.82/0.117 = 32.6 Liters which is 32600 milliliters which is unreasonable; check answer if you can
