Some conversion required. (mmolar into mol, or moles into mmol )
Molarity = moles of solute/Liters of solution
100 millimolar = 0.1 M glycine
Molarity = moles of solute/Liters of solution
manipulate algebraically
Liters of solution = moles of solute/Molarity
0.005 mole glycine/0.1 M glycine
= 0.05 Liters ( 1000 ml/1 L)
= 50 milliliters of solution
--------------------------------
The molarity is 0.718 moles/litre.
The answer is 0,625 moles.
The molarity is 0,025.
.487 Kl
Moles of solute/ kg of solvent = 2 molal solution
The molarity is 0.718 moles/litre.
The answer is 0,625 moles.
The molarity is 0,025.
.487 Kl
Moles of solute/ kg of solvent = 2 molal solution
Molarity is moles of solvent divided by liters of solution, so 6.42 / 1.75 = 3.67M.
3.33 degrees Celsius is the freezing point of a solution that contains 0.550 moles of Nal in 615 g of water.
The freezing point of a solution that contains 0.550 moles of NaI in 615 g of water is -3.33 degrees Celsius.
0.0296 M solution means 1000 ml contains 0.0296 moles of LiI By applying unitary method we get, no. of moles of LiI in 258.6ml =.00765456 moles
This molarity is 1,59.
Molarity = moles of solute/Liters of solution Molarity = 0.202 moles KCl/7.98 Liters = 0.253 M KCl solution ================
5.50eq noob