C = 4.83 m LiCl
mole ratio = X
moles = n
A = solute
B = solvent
XA = nA___
nA + nB
solution of LiCl in water
solute = LiCl
solvent = water = H2O
molality = moles of solute = moles of solute
kg of solvent 1.0 kg of solvent
Assume 1.000 kg H2O
1.000 kg H2O * 1000 g * 1 mol H2O = 55.49 mol H2O
1 kg 18.02 g H2O
XA = nA___
nA + nB
Solve for nA
XA( nA + nB ) = nA
XAnA = XBnB = nA
XAnB = nA - XAnA
XAnB = nA( 1 - XA)
XAnB___ = nA
( 1 - XA)
Plug in known amounts
nA = 0.08 * 55.49 mol H2O = 4.83 mol LiCl
1 - 0.08
If molality = moles of solute
1.0 kg of solvent
THEN molality of LiCl = 4.83 mol LiCl = 4.83 m
1.000 kg H2O
55.11g. The molecular weight of LiCl is 6.94 + 35.45 = 42.39 To make 1 liter of 0.65M we need 0.65 X 42.39 g = 27.5535g So to make 2 liters we need twice as much:- 55.11g
LiCl+H2SO4=LiHSO4+HCl
Li+
boiling point:1382 C Melting point: 605 C
2 Na + 2 LiCl = Li2 (Lithium) + 2NaCl (Table Salt)
You know because of solubility rules that LiCl disassociates 100% in water. So, knowing that molarity is equal to moles/liters Molarity LiCl = 1.97mol / 33.2 L Molarity = 0.059 M LiCl
To find the molarity, first calculate the number of moles of LiCl in 230 mL of water. Then, divide the moles of LiCl by the volume of water in liters (230 mL = 0.23 L) to get the molarity. In this case, 2.60 moles of LiCl in 0.23 L of water would result in a molarity of 11.30 mol/L.
To calculate the molarity, first convert the mass of LiCl to moles using its molar mass (6.94 g/mol for Li, 35.45 g/mol for Cl). Then, divide the moles of LiCl by the volume in liters (930 mL = 0.93 L) to get the molarity. The molarity of the solution would be around 5.2 mol/L.
To find the molarity, first calculate the number of moles of LiCl in 61.7 g using the molar mass of LiCl (42.39 g/mol). Then, divide the moles of LiCl by the volume of the solution in liters (0.250 L) to get the molarity.
It is a mixture of a salt compound (e.g. NaCl, NaCO2, LiCl, LiCO3, etc.) and water (or another fluid).
A mixture is a sample of matter that can be separated into different substances by physical means. This can include techniques such as filtration, distillation, or chromatography to isolate the individual components.
When lithium reacts with hydrochloric acid, it forms lithium chloride salt and hydrogen gas. The reaction is quite vigorous due to the reactivity of lithium with acids, so caution is needed when conducting this experiment.
To make a 4M solution in 20 ml, you would need 0.32 grams of LiCl. This can be calculated using the formula: moles = molarity x volume (in L), then converting moles to grams using the molar mass of LiCl.
Yes, this is an acid-base reaction. LiOH (strong base) and HCl (strong acid) react to form LiCl (salt) and H2O (water).
The equation for lithium chloride (LiCl) dissolving in water is LiCl(s) + H2O(l) -> Li+(aq) + Cl-(aq). This reaction shows the dissociation of LiCl into lithium ions (Li+) and chloride ions (Cl-) in aqueous solution.
The boiling point of 2 m KF in water is 102.4ºC. The boiling point of a 0.5 m aqueous solution of LiOH is the same as the boiling point of a 0.5 m aqueous solution of LiCl.
Sodium chloride and lithium chloride are very soluble in water.