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

Those answers look good to me. Good work.

To calculate each of the quantities, we'll use the following formulas:

(a) Grams of solute = Volume of solution (in L) x Molarity x Molar mass
(b) Molarity = Moles of solute / Volume of solution (in L)
(c) Moles of solute = Volume of solution (in L) x Molarity

Let's calculate each quantity step-by-step:

(a) Grams of solute in 185.8 mL of 0.267 M calcium acetate:
First, we need to convert mL to liters:
185.8 mL = 185.8/1000 L = 0.1858 L

Next, we can use the formula:
Grams of solute = Volume of solution x Molarity x Molar mass
Grams of solute = 0.1858 L x 0.267 mol/L x 7.85 g/mol = 0.401 g

Therefore, there are 0.401 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, we need to convert mL to liters:
500 mL = 500/1000 L = 0.5 L

Next, we can use the formula:
Molarity = Moles of solute / Volume of solution
Molarity = (21.1 g / )(1 mol/166.0 g) / 0.5 L = 0.254 M

Therefore, the molarity of the solution is 0.254 M.

(c) Moles of solute in 145.6 L of 0.850 M sodium cyanide:
We can use the formula:
Moles of solute = Volume of solution x Molarity
Moles of solute = 145.6 L x 0.850 mol/L = 123.76 mol

Therefore, there are 123.76 moles of solute in 145.6 L of 0.850 M sodium cyanide.

Sure! Let's calculate each of the quantities step by step.

(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 volume of the solution (in liters) by the molarity (mol/L), and then multiply it by the molar mass (g/mol) of calcium acetate.

First, we convert the volume from milliliters to liters:
185.8 mL = 185.8/1000 = 0.1858 L

Next, we use the formula: number of moles = molarity × volume
Number of moles of calcium acetate = 0.267 M × 0.1858 L = 0.0496 moles

Finally, we calculate the grams using the molar mass of calcium acetate:
Grams = moles × molar mass
Grams = 0.0496 moles × 158.17 g/mol (molar mass of calcium acetate)
Grams ≈ 7.85 g

Therefore, the grams of solute in 185.8 mL of 0.267 M calcium acetate is approximately 7.85 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 of the solution (in liters).

First, we convert the volume from milliliters to liters:
500 mL = 500/1000 = 0.5 L

Next, we calculate the moles of potassium iodine using the formula: moles = mass/molar mass.
Moles of potassium iodine = 21.1 g/166 g/mol (molar mass of KI) ≈ 0.127 moles

Finally, we calculate the molarity using the formula: molarity = moles/volume
Molarity ≈ 0.127 moles / 0.5 L ≈ 0.254 M

Therefore, the molarity of 500 mL of solution containing 21.1g of potassium iodine is approximately 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 molarity (mol/L) by the volume of the solution (in liters).

Moles of sodium cyanide = 0.850 M × 145.6 L = 123.76 moles

Therefore, the moles of solute in 145.6 L of 0.850 M sodium cyanide is approximately 124 mol.