Question

Determine the molarity for the following solutions.

2.0 moles of potassium iodide are dissolved into 0.10 L.
20. M

10.0 grams of potassium iodide are dissolved into 100 mL.
.

0.427 moles of aluminum carbonate are dissolved into 0.855 L of water.


2340 grams of aluminum carbonate are dissolved to make 1500 mL of solution.


Calculate the asked for quantity in the following solutions.

How many grams of potassium iodide will need to be dissolved to make a 100 mL of a 2.0 M solution?



You need to make a 0.5 M solution with 25.0 grams of potassium iodide. How much water do you need to dissolve it?


How many L do you need to dissolve 10.0 grams of aluminum carbonate into a 0.3 M solution?


How many grams of aluminum carbonate do you need to dissolve to make 8.0 L of a 0.75 M solution?

Answer 1

The molarity, amount and volumes required are:

• molarity = 20.0 M
• Molarity = 0.6 M
• Molarity = 0.5 M
• Molarity= 17.9 M
• mass of potassium iodide = 33.2 g
• Volume of water needed = 0.30 L or 300 mL
• Volume of water needed = 0.383 L or 383 mL

What is molarity of a solution?

Molarity of a solution is defined as the number of moles of a solute dissolved in a given volume in litres of the solution.

• Molarity of a solution = number of moles/volume in L
• Number of moles = mass/molar mass

Molarity of 2.0 moles of potassium iodide are dissolved into 0.10 L.

molarity = 2/0.1

molarity = 20. M

Molarityof 10 grams of potassium iodide dissolved into 100 mL

molarity mass of potassium iodide = 166 g/mol

100 mL = 0.1 L

number of moles = 10/166 = 0.06 moles

Molarity = 0.06/ 0.1

Molarity = 0.6 M

.

Molarityof 0.427 moles of aluminum carbonate dissolved into 0.855 L of water.

Molarity = 0.427/0.855

Molarity = 0.50 M

Molarity of 2340 grams of aluminum carbonate dissolved to make 1500 mL of solution.

1500 mL = 1.5 L

molar mass of aluminium carbonate = 87 g/mol

number of moles = 2340/87 = 26.9 moles

Molarity= 26.9/1.5

Molarity = 17.9 M

How many grams of potassium iodide will need to be dissolved to make a 100 mL of a 2.0 M solution?

100 mL = 0.1 L

molarity = 2.0

number of moles = 2.0 × 0.1 = 0.2 moles

molar mass of potassium iodide = 166 g/mol

mass = number of moles × molar mass

mass = 0.2 × 166

mass of potassium iodide = 33.2 g

How much water is required to make a 0.5 M solution with 25.0 grams of potassium iodide?

Number of moles of potassium iodide = 25/166 = 0.15 moles.

Volume = number of moles/molarity

Volume of water needed = 0.15/0.6

Volume of water needed = 0.30 L or 300 mL

How many L do you need to dissolve 10.0 grams of aluminum carbonate into a 0.3 M solution?

Number of moles of aluminium carbonate = 10/87 = 0.115 moles.

Volume = number of moles/molarity

Volume of water needed = 0.115/0.3

Volume of water needed = 0.383 L or 383 mL

Therefore, the molarity, amount and volumes required are:

• molarity = 20.0 M
• Molarity = 0.6 M
• Molarity = 0.427/0.855
• Molarity= 17.9 M
• mass of potassium iodide = 33.2 g
• Volume of water needed = 0.30 L or 300 mL
• Volume of water needed = 0.383 L or 383 mL

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