4 mol over 0.800 kg
Molarity = moles of solute/Liters of solution
So, we need moles of glucose.
4 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.02220 moles glucose
---------------------------------------------------------------------------------------------------
Molarity = 0.02220 moles C6H12O6/1 Liter
= 0.02 M glucose
----------------------
The molarity of a solution is 0.400 mol divided by 4.00 kg which equals 0.100 mol/kg solvent which equals 0.100 m.
The molality of a solution is the moles of solute divided by the mass of the solvent (in kg):
molality = mass/kg So, molality = 4.0 moles / 3.2 kg = 1.25 moles/kg
The molality of this solution is 1,905; to calculate the molarity you need the density of the solution.
4 mol/0.800 kg
The molality is 5.
The molality is 5.
0.1 N KCl is the same as 0.1 M KCl. This requires one to dissolve 0.1 moles per each liter of solution. The molar mass of KCl is 74.6 g/mol. So 0.1 moles = 7.46 gDissolve 7.46 g KCl in enough water to make 1 liter (1000 ml)Dissolve 3.73 g KCl in enough water to make 0.5 liter (500 ml)Dissolve 0.746 g KCl in enough water to make 0.1 liter (100 ml)etc., etc.
The molarity is 0,388.
KCl is soluble in DMF
online wha u think?
To decrease melting point of NaCl
Dissolve 566.25g of KCl in 3L.
the solution of KCl looks like a clear water with little tint of gray swirls
18%
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
4.0%
32 g KCl
KCl and CCl4 do they form solution
Minimum 102,6 g of KCl.
moles KCL = ( M solution ) ( L of solution )moles KCl = ( 0.83 mol KCl / L ) ( 1.7 L ) = 1.41 moles KCl
Weigh 22.35 grams of KCl and Dissolve in 100 mL of Distilled Water
Potassium chloride, KCl, is a salt; it dissolves in water and would be considered a solute when it does so.
This is a solution of 10 g KCl/100 g water.