Given: 0.5 g NaCl; 0.05 L of solution.
1) Find the molar mass of NaCl.
(22.99) + (35.45) = 58.44 g/mol of NaCl
2) Convert grams of NaCl to moles of NaCl.
(0.5 g NaCl) X (1 mol NaCl / 58.44 g NaCl) = 0.00855578 mol NaCl
3) Use the molarity equation, M = mol / L, to solve for molarity (M).
(0.00855578 mol NaCl / 0.05 L) = 0.17 M
What is the molarity of a solution that has 50 grams of KOH dissolved in enough water to make 800 mL of solution?
0.5 divided by 1.5 is 0.33333 grams per litre. The molecular weight of sodium is 23g/mol. 0.33333/23 is 0.0145 molar.
Use the molarity equation, M = mol / L.
0.5 mol / 0.05 L = 10M
0.0004 M
.800 M
0.583 M
12
The molecular weight of NaCl is 58.443 g/mol. 78/58.443 is 1.300412367606 moles. It is in litres so it is 1.300412367606 molar.
4m
Molarity = moles of solute/volume of solution Find moles NaCl 55 grams NaCl (1mol NaCl/58.44 grams) = 0.941 moles NaCl Molarity = 0.941 moles NaCl/35 Liters = 0.027 Molarity NaCl ( sounds reasonable as 55 grams is not much in 35 Liters of water, which would be about 17.5 2 liter sodas )
Because you have 6.68 moles of Li2SO4 and 2.500 liters of water, the overall molarity of your solution is 2.67 M.
If we add 1 L of water to 1 L of 2 M NaCl solution will give 2 L of 1 M NaCl solution
1.3g
The molecular weight of NaCl is 58.443 g/mol. 78/58.443 is 1.300412367606 moles. It is in litres so it is 1.300412367606 molar.
In moles of substance dissolved in 1 L of water -Apex
4m
Molarity = moles of solute/volume of solution Find moles NaCl 55 grams NaCl (1mol NaCl/58.44 grams) = 0.941 moles NaCl Molarity = 0.941 moles NaCl/35 Liters = 0.027 Molarity NaCl ( sounds reasonable as 55 grams is not much in 35 Liters of water, which would be about 17.5 2 liter sodas )
Because you have 6.68 moles of Li2SO4 and 2.500 liters of water, the overall molarity of your solution is 2.67 M.
[117(g NaCl) / 58.5(g NaCl/mol NaCl)] / 40.0(L solution) = [117/58.5]/40.0 = 2.00(mol NaCl) / 40.0(L) = 0.0500 mol NaCl / L solution = 0.0500 M
Divide grams (mass) by molar mass to find moles58.44 (g NaCl/L) / [22.99+35.45](g NaCl/mol NaCl)= 1.000 mol/L NaCl
when any ionizable cpmpound is dissolved in water it becomes an electrolytic solution as NaCl or HCl in water....
When you take a solution and make it pure, thus removing the solvent and returning the dissolved compound to its pure molarity
When you take a solution and make it pure, thus removing the solvent and returning the dissolved compound to its pure molarity
Molarity=moles of solute/liters of solvent=5mol/10L=0.5M