Use c1*V1=c2*V2 to calculate:
(this goes for ANY molarity, not only for acetic acid as questioned)
5(M) * 2(L) = x(M) * 7.5(L) , so x = molarity of the diluted = 1.3 M
The resulting concentration in M is 0,0118 (approx. 0,012).
initial molarity*initial volume= final molarity*final volume Initial molarity= 1.50M Initial volume= 20.00ml Final Volume=150.0ml Thus final molarity =1.50M*20ml/150ml=0.200M. New molar concentration= final molarity
1.0 molal
Yes. The volume you have of a particular solution does not have anything to do with the concentration of that solution.
Usually not, but you can add more solvent and remove a compensating volume of the solute. This is easest if it is a liquid-in-liquid or a gas-in-gas solution.
When a given solution is diluted, it concentration is usually lowered as well. The density of the given solution also changes when the solution is diluted.
The concentration is 34,32 g/L.
.33
The concentration of a solution is basically how strong the solution is.
The resulting concentration in M is 0,0118 (approx. 0,012).
The concentration of a solution is typically given in terms of the volume of solution, in liters.
The concentration of the diluted solution will be 15(300/1000) = 4.5 %, if the percent is expressed on a weight/volume basis.
If the concentration of alcohol and water solution is 25 percent alcohol by volume, the volume of alcohol in a 200 solution is 50.
initial molarity*initial volume= final molarity*final volume Initial molarity= 1.50M Initial volume= 20.00ml Final Volume=150.0ml Thus final molarity =1.50M*20ml/150ml=0.200M. New molar concentration= final molarity
Are you sure you mean 0.211 m not 0.211 M Is the concentration molality or molarity? If 55 ml of a 0.211 m NaOH is diluted to a final volume of 125 ml what is the concentration of NaOH in the diluted solution?NaOH not NaHO the compound is Sodium hydroxide Molar mass of NaOH = 23 + 16+ 1 = 40 0.211 moles of NaOH = 0.211 * 40 = 8.44 grams of NaOH per liter of solution 55 ml = 0.055 Liter 8.44 grams of NaOH per liter of solution * 0.055 liter = 0.4642 g of NaOH Moles of NaOH = 8.44 ÷ 40 = 0.211 moles 0.211 moles of NaOH in 125 ml = 0.211 moles ÷ 0.125 L = 1.688 m
ask your chemistry teacher at 12 PM tomorrow
It can be diluted by adding the appropriate amount of water to the solution.