calculate the molarity of each of the following solutions. Help!!

(a) 0.36 mol of LiNO3 L of solution

(b) 70.7 g C2H6O in 2.25 L of solution

(c) 11.56 mg KI in 114.2 mL of solution

M = moles/L

the first one (a) I was suppose to write 6.36L of LiNO3 soooooo but ur M=moles/L

I take 0.36/6.36L and get 0.0566 and i'm suppose to give the answer in 2 significant figures which I think is 0.05 but it I got it wrong.......the anwser is

M= 5.5x10-2 M

what did i do wrong???

Sure! I can help you calculate the molarity of each of the solutions. Remember, the formula to calculate molarity is:

Molarity (M) = moles of solute / volume of solution in liters

Let's calculate the molarity for each solution:

(a) 0.36 mol of LiNO3 in L of solution:

Molarity (M) = 0.36 mol / L of solution

Since the volume of the solution is not specified, we can't determine the molarity accurately without knowing the volume. Please provide the volume of the solution in order to calculate the molarity accurately.

(b) 70.7 g C2H6O in 2.25 L of solution:

First, we need to convert the mass of C2H6O to moles:

Molar mass of C2H6O = 2* 12.01 g/mol + 6 * 1.01 g/mol + 16 * 1.01 g/mol = 46.07 g/mol

Number of moles of C2H6O = 70.7 g / 46.07 g/mol

Molarity (M) = Number of moles of C2H6O / Volume of solution in liters

Molarity (M) = (70.7 g / 46.07 g/mol) / 2.25 L

(c) 11.56 mg KI in 114.2 mL of solution:

First, let's convert the mass of KI to moles:

Molar mass of KI = 39.09 g/mol + 126.9 g/mol = 166.99 g/mol

Number of moles of KI = 11.56 mg / 166.99 g/mol

Next, we need to convert the volume of the solution to liters:

Volume of solution = 114.2 mL = 0.1142 L

Molarity (M) = Number of moles of KI / Volume of solution in liters

Molarity (M) = (11.56 mg / 166.99 g/mol) / 0.1142 L

Please note that the final molarity values will depend on the accurate values of the volume of the solution.

To calculate the molarity of a solution, you need to know the moles of solute and the volume of solution in liters. Molarity is defined as the moles of solute divided by the liters of solution.

Let's calculate the molarity for each of the given solutions:

(a) 0.36 mol of LiNO3 in L of solution:
Molarity = moles of solute / volume of solution

Moles of solute = 0.36 mol
Volume of solution = 1 L

Molarity = 0.36 mol / 1 L = 0.36 M

Therefore, the molarity of the solution is 0.36 M.

(b) 70.7 g C2H6O in 2.25 L of solution:
First, we need to convert grams of C2H6O to moles using its molar mass.
The molar mass of C2H6O (ethyl alcohol) is:
(2 * atomic mass of C) + (6 * atomic mass of H) + (1 * atomic mass of O) = 46.07 g/mol

Moles of C2H6O = Mass of C2H6O / Molar mass of C2H6O
= 70.7 g / 46.07 g/mol

Now, we have moles of solute, and the volume of solution is given as 2.25 L.

Molarity = moles of solute / volume of solution
= (70.7 g / 46.07 g/mol) / 2.25 L

Therefore, the molarity of the solution is the final calculated value.

(c) 11.56 mg KI in 114.2 mL of solution:
First, we need to convert milligrams of KI to grams.

Mass of KI = 11.56 mg = 0.01156 g
Moles of KI = Mass of KI / Molar mass of KI

The molar mass of KI (Potassium iodide) is:
(1 * atomic mass of K) + (1 * atomic mass of I) = 166 g/mol

Moles of KI = 0.01156 g / 166 g/mol

Now, we have moles of solute, and the volume of solution is given as 114.2 mL.

To calculate molarity, we need to convert milliliters to liters:

Volume of solution = 114.2 mL = 0.1142 L

Molarity = moles of solute / volume of solution
= (0.01156 g / 166 g/mol) / 0.1142 L

Therefore, the molarity of the solution is the final calculated value.