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Chapter 12: Problem 202

What is the molarity of \(3.69 \mathrm{~mL}\) of solution containing \(0.0025\) mole of calcium chloride?

Short Answer

Expert verified
The molarity of the calcium chloride solution is approximately \(0.6778 \mathrm{~M}\).

Step by step solution

01

Convert the volume from milliliters to liters

To convert the volume of the solution from milliliters (mL) to liters (L), we need to use the conversion factor: 1 L = 1000 mL. Volume in liters = volume in milliliters ÷ 1000 Volume in liters = 3.69 mL ÷ 1000
02

Calculate the molarity

Now, apply the formula for molarity: Molarity (M) = moles of solute ÷ volume of solution in liters Moles of solute (CaCl₂) = 0.0025 moles Volume of solution in liters = 0.00369 L (from Step 1) Molarity = 0.0025 moles ÷ 0.00369 L
03

Compute the molarity

Now, simply divide the moles of solute by the volume of the solution in liters to find the molarity: Molarity = 0.0025 moles ÷ 0.00369 L ≈ 0.6778 M Therefore, the molarity of the calcium chloride solution is approximately 0.6778 M.

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Most popular questions from this chapter

Give precise instructions to your laboratory assistant as to how to prepare \(1.00 \mathrm{~L}\) of a \(0.250 \mathrm{M}\) aqueous solution of sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\). Remember that she will be measuring out the sucrose in grams. She has available a 1-Lvolumetric flask.

A \(50.00\) -mL sample of hydrochloric acid of unknown concentration was neutralized by \(47.35 \mathrm{~mL}\) of \(0.01020 \mathrm{M}\) sodium hydroxide solution. Calculate the concentration of the original HCl solution.

An alcohol-water solution is \(35.00\) vol \% alcohol. How much solution is required to obtain \(200.0\) \(\mathrm{mL}\) of alcohol?

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A bottle contains \(1.00 \mathrm{~L}\) of a stock solution of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) of unknown concentration. A lab technician dilutes \(5.00 \mathrm{~mL}\) of the stock solution to \(100.0 \mathrm{~mL}\) with water. He then determines the \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) concentration of this solution to be \(0.0478 \mathrm{M}\). What is the \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) concentration of the stock solution?
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