Acids and Bases

012

Molarity = moles of solute/Liters of solution ( 230 ml = 0.0230 Liters) Molarity = 2.60 moles LiCl/0.230 Liters = 11. 3 M LiCl -----------------

Molarity = moles of solute/Liters of solution Molarity = 25 moles sucrose/50 liters H2O = 0.5 M sucrose

since molarity = (number of moles) / (volume of solution), the molarity of a solution with 2.3 moles of sodium chloride in 0.45 liters of water will be 5.11 moles.

Molarity = moles of solute/Liters of solution Molarity = 6 Moles NaCl/2 Liters = 3 M NaCl ========

1620 ml = 1.62 Liters Molarity = moles of solute/Liters of solution Molarity = 2 moles NaOH/1.62 Liters = 1 M NaOH solution ===============

MgCl2 Molarity = moles of solute/Liters of solution ( 250 ml = 0.250 L ) Get moles MgCl2 80 grams MgCl2 (1 mole MgCl2/95.21 grams) = 0.8402 moles MgCl2 Molarity = 0.8402 moles MgCl2/0.250 Liters = 3.4 M MgCl2 ----------------

NaF has an atomic mass of 42, therefore if you dissolve 42 grams (1 mole) of it in 2 liters of water you get a molarity of 0.5 moles/Liter.

This molarity is 3 M.

Molarity means number of moles in a solute volume. So molarity is 2moldm-3.

Because you have 6.68 moles of Li2SO4 and 2.500 liters of water, the overall molarity of your solution is 2.67 M.

Concentration of a solution is recorded in molarity (M). Molarity is the moles of solute divided my liters of solution. So to find the concentration of a solution, calculate the number of moles of the solute (the chemical being dissolved) and measure the number of liters of the solution (the water), then divide them.

Molarity = moles of solute/Liters of solutionSo, get moles HCl.73 grams HCl (1 mole HCl/36.458)= 2.00 moles HCl---------------------------Molarity = 2.00 moles HCl/2 Liters= 1 M HCl=======

This is just easy. MOLARITY OF THIS COMPOUND IS 3MOLDM-3.

Molarity (concentration ) = moles of solute/Liters of solution 250.0 ml = 0.250 liters 2.431 grams H2C2O4 * 2H2O ( 1mole cpd/ 126.068 grams) = 0.01928 moles H2C2O4 * 2H2O Molarity = 0.01928 moles cpd/0.250 liters = 0.07712 Molarity

Molarity = moles of solute/Liters of solution ( get moles of HNO3 and 300 ml = 0.300 Liters ) 0.31 grams Nitric acid (1 mole HNO3/63.018 grams) = 0.004919 moles HNO3 Molarity = 0.004919 moles HNO3/0.300 Liters = 0.0164 M HNO3

This molarity is 3 mol/L.

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 )

Molarity= moles/Liters To change grams to moles you divide by the mole weight which is listed on the periodic table. Mol= grams/mol weight The Mole weight of Magnesium Chloride is 59.8 grams/mol Mol=128g/59.8 Mol=2.14 Now, you put the number of moles and Liters into the equation Molarity=2.14 mol/1L Molarity=2.14 So, the molarity is 2.14 M

Molarity = number of moles / number of liters. For this question the number of moles is 3 and the number of liters is 0.5 So 3/0.5 = 6 The solution is 6 M.

2 moles in 0.5 liters will mean 4 moles in 1 liter. So the molarity is 4 molar.

By it's definition, molarity means how many moles are in a litre. As you already have your moles and it is per litre already, the answer is 2 molar.

6 kg = 6000 grams and density of water = 1.00 grams/milliliters. 1.00 g/ml = 6000 grams/X ml = 6000 ml which = 6 liters ======================== Molarity = moles of solute/Liters of solution Molarity = 2 moles NaOH/6 Liters = 0.3 M NaOH solution -----------------------------

The molarity is figured out by the number of moles of NaOH over the number of liters of water. so if the molarity is 0.5M and there is 1.0L of water then there has to be 0.5 moles of NaOH in the solution.

Get moles sodium sulfate first.15.5 grams Na2SO4 (1 mole Na2SO4/142.05 grams) = 0.1091 moles Na2SO4Now,Molarity = moles of solute/ Liters of solution ( 35 ml = 0.035 Liters )Molarity = 0.1091 moles Na2SO4/0.035 Liters= 3.1 M Na2SO4 solution==================

Find moles potassium iodide first.2.41 grams KI (1 mole KI/166 grams) = 0.01452 moles KIMolarity = moles of solute/Liters of solution ( 100 ml = 0.1 Liters )Molarity = 0.01452 moles KI/0.1 Liters= 0.145 M KI solution================

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