Calculate each of the following quantities:
(a) Grams of solute in 185.8 mL of 0.267 M calcium acetate 7.85g Ca(C2H3O2)2
(b) Molarity of 500. mL of solution containing 21.1g of potassium iodine 0.254 M KI
(c) Moles of solute in 145.6 L of 0.850 M sodium cyanide 124 mol NaCN
To calculate each of the quantities, we will use the formulas:
(a) Grams of solute = Volume of Solution (in liters) x Molarity x Molar Mass
(b) Molarity = Grams of solute / Volume of Solution (in liters)
(c) Moles of solute = Volume of Solution (in liters) x Molarity
Let's calculate each quantity one by one:
(a) Grams of solute in 185.8 mL of 0.267 M calcium acetate:
First, convert the volume from mL to L:
185.8 mL = 185.8 / 1000 = 0.1858 L
Now, use the formula:
Grams of solute = Volume of Solution x Molarity x Molar Mass
Grams of solute = 0.1858 L x 0.267 M x 7.85 g/mol = 0.4047 g
Therefore, there are 0.4047 grams of solute in 185.8 mL of 0.267 M calcium acetate.
(b) Molarity of 500 mL of solution containing 21.1 g of potassium iodine:
First, convert the volume from mL to L:
500 mL = 500 / 1000 = 0.5 L
Now, use the formula:
Molarity = Grams of solute / Volume of Solution
Molarity = 21.1 g / 0.5 L = 42.2 M
Therefore, the molarity of the solution is 42.2 M.
(c) Moles of solute in 145.6 L of 0.850 M sodium cyanide:
Use the formula:
Moles of solute = Volume of Solution x Molarity
Moles of solute = 145.6 L x 0.850 M = 123.76 mol
Therefore, there are 123.76 moles of solute in 145.6 L of 0.850 M sodium cyanide.
To calculate each of the quantities, we need to use the formula:
Quantity = Concentration x Volume
(a) Grams of solute in 185.8 mL of 0.267 M calcium acetate:
To calculate the grams of solute, we need to multiply the concentration (0.267 M) by the volume (185.8 mL). However, since the molar mass of calcium acetate (Ca(C2H3O2)2) is also provided, we need to perform an additional step to convert moles to grams.
First, let's calculate the moles of calcium acetate:
moles = concentration x volume
moles = 0.267 M x 0.1858 L
moles = 0.0496 mol
Now, let's convert moles to grams using the molar mass of calcium acetate:
grams = moles x molar mass
grams = 0.0496 mol x 158.17 g/mol
grams = 7.84 g
Therefore, the grams of solute in 185.8 mL of 0.267 M calcium acetate is approximately 7.84 g.
(b) Molarity of 500 mL of solution containing 21.1g of potassium iodine:
To calculate the molarity, we need to divide the moles of solute by the volume in liters.
First, let's calculate the moles of potassium iodine:
moles = grams / molar mass
moles = 21.1 g / 166.0028 g/mol
moles = 0.1273 mol
Now, let's calculate the molarity:
Molarity = moles / volume (in liters)
Molarity = 0.1273 mol / 0.5 L
Molarity = 0.254 M
Therefore, the molarity of 500 mL of solution containing 21.1g of potassium iodine is 0.254 M.
(c) Moles of solute in 145.6 L of 0.850 M sodium cyanide:
To calculate the moles of solute, we need to multiply the concentration (0.850 M) by the volume (145.6 L).
moles = concentration x volume
moles = 0.850 M x 145.6 L
moles = 123.76 mol
Therefore, the moles of solute in 145.6 L of 0.850 M sodium cyanide is approximately 123.76 mol.