Sodium and chloride have fixed oxidation states. Sodium is Na+ and Chloride is Cl-. Their ratio is 1:1 (1 to 1). So in this particular compound, there is only one sodium ion and 1 x 2.5 = 2.5 moles (answer).
Use: quantity in mole = volume * concentration. Make units equalling.
50 (mL) * 2.15 (mol/L) = 50 (mL) * 2.15 (mmol/mL) = 50*2.15 (mL * mmol / mL) = 107.5 mmol = 0.108 mol NaCl
5.0 (L) * 0.100 (mol/L) = 0.50 molNaCl
0,45 L of 2,5 M sodium chloride solution contain 65,7 g salt.
1.65
42.5 grams
2
0.5
There is twice the change in colligative properties in the sodium chloride solution than in the glucose solution.
The equation for the reaction: Na3PO4 + 3KCl ---> 3NaCl + K3PO4 Shows that 1 mol of sodium phoshpate will yield 3 mol of sodium chloride. The number of moles of Na3PO4 present in the solution in this reaction = concentration x volume => 0.0250 mol.dm-3 x 0.03 dm3 = 0.00075 mol. When reacted with excess KCl, this will give 0.00075 mol x 3 = 0.00225 mol of NaCl. The mass of NaCl formed = mol x RFM => 0.00225 x 58.5 = 0.132 g
This method uses a back titration with potassium thiocyanate to determine the concentration of chloride ions in a solution.Before the titration an excess volume of a standardized silver nitrate solution is added to the solution containing chloride ions, forming a precipitate of silver chloride (AgCl). The term 'excess' is used as the moles of silver nitrate added are known to exceed the moles of sodium chloride present in the sample so that all the chloride ions present will react.Ag+ + Cl- AgCl(s) (Ksp = 1.70 × 10−10)Excess WhiteExcess of Ag+ is back titrated with SCN-.
Molarity is calculated as moles of solute divided by volume of solution in liters. In this case, you have 2 moles of sodium chloride in a 0.5 liter solution. So the molarity would be 2 moles / 0.5 L = 4 M.
321
This depends on: - the volume of the drop - the concentration of sodium chloride solution
Density=Mass/Volume
It depends on the volume, if we consider 1 liter of the solution 500 mg of sodium chloride is needed.
1. A volume of powdered sodium chloride.2. A volume of water solution of sodium chloride with a non-specified concentration in this case..
Molarity is moles/liter, so in order to find the moles of a substance in a given volume, simply multiply molarity with volume (in liters). n=M*V
Sodium chloride solution, dextrose solution, ringer's solution and lactated ringer's solution are all common large volume parenteral products.
There is twice the change in colligative properties in the sodium chloride solution than in the glucose solution.
What is the volume of 35.7g of sodium chloride in 100cm3 of cold water?
The volume is 0,3 mL.
The answer is 48,17 L.
The total volume is a little decreased.
You need 116,88 g dried and pure sodium chloride.