Molarity = moles of solute/Liters of solution ( 50 ml = 0.05 Liters )
12 M HCl = moles HCl/0.05 Liters
= 0.60 moles HCl
22 g HCl
Molarity = moles of solute/Liters of solution (40 ml = 0.04 Liters) algebraically manipulated, Moles of solute = Liters of solution * Molarity Moles HCl = (0.04 Liters)(0.035 M) = 0.0014 moles HCl ==============
Molarity = moles of solute/liters of solution, so... 0.400M HCl (X mols HCl/0.250L ) = 0.100 moles HCl
Molarity = moles of solute/liters of solution Molarity = 0.597 moles HCl/0.169 liters = 3.53 M HCl ------------------
Molarity = moles of solute/Liters of solutionSo, get moles HCl.73 grams HCl (1 mole HCl/36.458)= 2.00 moles HCl---------------------------Molarity = 2.00 moles HCl/2 Liters= 1 M HCl=======
First.Molarity = moles of solute/Liters of solution0.22 M HCl = X moles/1.0 L= 0.22 moles HCl--------------------------Second.0.22 moles HCl (36.458 grams/1 mole HCl)= 8.0 grams hydrochloric acid needed===========================
Molarity = moles of solute/Liters of solution (40 ml = 0.04 Liters) algebraically manipulated, Moles of solute = Liters of solution * Molarity Moles HCl = (0.04 Liters)(0.035 M) = 0.0014 moles HCl ==============
Molarity = moles of solute/liters of solution, so... 0.400M HCl (X mols HCl/0.250L ) = 0.100 moles HCl
Molarity = moles of solute/liters of solution Molarity = 0.597 moles HCl/0.169 liters = 3.53 M HCl ------------------
Molarity = moles of solute/Liters of solutionOr, for our purposes....,Liters of solution (volume) = moles of solute/MolarityVolume (liters) = 0.150 moles HCl/4.00 M HCl= 0.0375 liters = 37.5 milliliters======================
molarity = number of moles of solute / volume of solution number of moles of solute = molarity x volume of solution number of moles of solute = 12 (mole/L) x 15/1000 (L) number of moles of solute = 0.18 mole
Molarity = moles of solute/Liters of solutionSo, get moles HCl.73 grams HCl (1 mole HCl/36.458)= 2.00 moles HCl---------------------------Molarity = 2.00 moles HCl/2 Liters= 1 M HCl=======
First.Molarity = moles of solute/Liters of solution0.22 M HCl = X moles/1.0 L= 0.22 moles HCl--------------------------Second.0.22 moles HCl (36.458 grams/1 mole HCl)= 8.0 grams hydrochloric acid needed===========================
The number of moles of a solute will not change as a solution is diluted, however, the concentration of the solute will decrease. If you were to evaporate the water from the diluted solution, you would have the same number of moles of solute as when you started. You can test this by comparing the mass of the solute before producing the solution to the mass of the solute after the solution was diluted. The two masses should be the same.
We can use PV = nRT to find moles of HCl (1 atm)(4.60 L) = n(0.08206 L*atm/mol*K)(298.15 K) moles HCl = 0.188 moles Molarity = moles of solute/Liters of solution Molarity = 0.188 moles HCl/0.240 liters = 0.783 Molar HCl =============
1. First, remember definition of M (moles), M = moles of species / L. 0.33 M = 0.33 moles HCl / L 2. Then, multiple your volume by the molar concentration: 0.33 moles HCl / L x 0.70 L = 0.231 moles HCl or you can say n=CONCENTRATION multiply by VOLUME(HCl) which gives 2310 mol HCl It's helpful to carry the units with your calculations. That way you can check that numerators and denominators cancel to give you the units of your answer.
Balanced equation. NaOH + HCl -> NaCl + H2O all one to one. find moles HCl. 11 grams HCl (1 mole HCl/36.458 grams ) = 0.3017 moles HCl Moles HCl same as moles NaOH Molarity = moles of solute/Liters of solution 1.06 M NaOH = 0.3017 moles NaOH/liters of solution = 0.2846 Liters this is equal to..... 285 milliliters of NaOH needed
Balanced equation first, last and always! Na2CO3 + 2HCl - > 2NaCl + CO2 + H2O 2.5 g Na2CO3 ( 1 mole Na2CO3/105.99 g)(2 mole HCl/1 mole Na2CO3) 0.04717 moles HCl -------------------------Now, Molarity = moles of solute/liters of solution or, for our purposes liters of solution = moles of solute/Molarity Liters HCl = 0.04717 moles HCl/0.60 M HCl = 0.0786 liters (1000 milliliters/1 liter) = 78. 6 milliliters HCl solution needed