because that tip of the calomel electrode is made up of porous nature..If we keep it in environment, its get oxidised,,
I just got done talking with a vendor for pH probes and he explained that the KCl solution helps facilitate the measurement of the hydrogen ions with the electrode. It will actually seep out in small quantities from the tip which is a membrane.
KCl and CCl4 do they form solution
Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed
18%
0.4
It consists of tube in the bottom of which is a layer of mercury ,over which is placed a paste of Hg+HgCl2.The remaining portion of cell is filled with a solution of normal or decinormal or saturated solution of KCl. a platinum wire dipping into the mercury layer is used for making electrical contact. the side tube is used for electrode is formulated as;Hg,Hg2Cl2,KCl. the electrode can be coupled with the hydrogen electrode of unknown pH.
I just got done talking with a vendor for pH probes and he explained that the KCl solution helps facilitate the measurement of the hydrogen ions with the electrode. It will actually seep out in small quantities from the tip which is a membrane.
moles KCl = ( M solution ) ( V solution in L )moles KCl = ( 2.2 mol KCl / L solution ) ( 0.635 L of solution )moles KCl = 1.397 moles KCl
KCl and CCl4 do they form solution
moles KCL = ( M solution ) ( L of solution )moles KCl = ( 0.83 mol KCl / L ) ( 1.7 L ) = 1.41 moles KCl
MW KCl = 74.6 g/mol2.39 gKCl * (1 mol KCL/74.6 g KCl)*(1 L solution/0.06 mol KCL) = 0.534 L
M= moles in solution/liters so plug in what you know 3.0M of KCl solution = moles in solution/ 2.0L multiply both sides by 2.0L moles solute = 1.5 moles KCl so you need 1.5 moles KCl to prepare the solution
Dissolve 566.25g of KCl in 3L.
I did not know that you could get a concentration of 75.66 M KCl, but; Molarity = moles of solute/Liters of solution 75.66 M KCl = moles KCl/1 liter = 75.66 moles of KCl 75.66 moles KCl (74.55 grams/1 mole KCl) = 5640 grams KCl that is about 13 pounds of KCl in 1 liter of solution. This is why I think there is something really wrong with this problem!
Molarity = moles of solute/Liters of solution 0.75 M KCl = moles KCl/2.25 Liters = 1.6875 moles KCl (74.55 grams/1 mole KCl) = 126 grams of KCl needed
0.745*0.5 g
Need mole KCl first. 4.88 grams KCl (1 mole KCl/74.55 grams) = 0.06546 moles KCl =======================now, Molarity = moles of solute/Liters of solution ( 423 ml = 0.423 Liters ) Molarity = 0.06546 moles KCl/0.423 Liters = 0.155 M KCl ------------------