I need help working the problems, I don't know how I got the right answers
Calculate each of the following quantities:
(a) Grams of solute in 185.8 mL of 0.267 M calcium acetate
(b) Molarity of 500. mL of solution containing 21.1g of potassium iodine
(c) Moles of solute in 145.6 L of 0.850 M sodium cyanide
I thought I worked this problem a day or so ago for you Emily.
You did... I only posted one question today, which was earlier
So why is this problem posted again? Was there something about my earlier response you didn't understand?
Sure, I can help you with that! Let's go through each of the problems one by one and I'll explain how to calculate the desired quantities.
(a) Grams of solute in 185.8 mL of 0.267 M calcium acetate:
To find the grams of solute, we first need to understand the meaning of molarity. Molarity (M) is defined as the number of moles of solute per liter of solution. In other words, it's the concentration of the solute in the solution.
Given:
Volume of solution (V) = 185.8 mL
Molarity (M) = 0.267 M
To calculate the grams of solute, we can use the following formula:
Grams of solute = Molarity × Volume × Molar mass
The molar mass of calcium acetate (Ca(CH3COO)2) can be found by adding up the atomic masses of its constituent elements:
Molar mass of Ca = 40.08 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol (hydrogen has 2 atoms in each acetate)
Molar mass of O = 16.00 g/mol (oxygen has 2 atoms in each acetate)
Molar mass of calcium acetate = (40.08 g/mol) + (2 * (12.01 g/mol) + 2 * (1.01 g/mol)) + (2 * 16.00 g/mol) = 40.08 g/mol + 24.02 g/mol + 2.02 g/mol + 32.00 g/mol = 98.12 g/mol
Now, we can plug the values into the formula:
Grams of solute = (0.267 M) × (185.8 mL) × (98.12 g/mol) / (1000 mL/1 L) = (0.267 M) × (0.1858 L) × (98.12 g/mol) = 4.953 g
Therefore, the grams of solute in 185.8 mL of 0.267 M calcium acetate is approximately 4.953 grams.
(b) Molarity of 500. mL of solution containing 21.1g of potassium iodine:
To find the molarity, we need the number of moles of solute and the volume of the solution.
Given:
Mass of solute = 21.1 g
Volume of solution (V) = 500 mL
First, we need to calculate the number of moles of potassium iodine (KI) using its molar mass:
Molar mass of K = 39.10 g/mol
Molar mass of I = 126.90 g/mol
Molar mass of potassium iodine = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol
Number of moles of KI = (21.1 g) / (166.00 g/mol) = 0.127 moles
Now, we can calculate the molarity using the formula:
Molarity = Moles / Volume
Molarity = (0.127 moles) / (500 mL/1000 mL/1 L) = 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 find the moles of solute, we need the molarity and the volume of the solution.
Given:
Molarity (M) = 0.850 M
Volume of solution (V) = 145.6 L
To calculate the moles of solute, we can use the formula:
Moles = Molarity × Volume
Moles = (0.850 M) × (145.6 L) = 123.86 moles
Therefore, the moles of solute in 145.6 L of 0.850 M sodium cyanide is approximately 123.86 moles.