Question Video: Calculating the Mass of Solute Needed to Prepare a Solution with a Desired Concentration and Volume | Nagwa Question Video: Calculating the Mass of Solute Needed to Prepare a Solution with a Desired Concentration and Volume | Nagwa

Question Video: Calculating the Mass of Solute Needed to Prepare a Solution with a Desired Concentration and Volume Chemistry • First Year of Secondary School

A student wants to prepare a 0.1 M solution of silver nitrate (AgNO₃) in a volumetric flask that can contain 100 mL of water. How much silver nitrate does the student need to dissolve? Give your answer to 1 decimal place. [N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol]

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Video Transcript

A student wants to prepare a 0.1-molar solution of silver nitrate, AgNO3, in a volumetric flask that can contain 100 milliliters of water. How much silver nitrate does the student need to dissolve? Give your answer to one decimal place.

The molar mass of nitrogen is 14 grams per mole, oxygen is 16 grams per mole, and silver is 108 grams per mole. In this question, a student is going to prepare a solution of silver nitrate in a volumetric flask that is designed to contain 100 milliliters of solution. The solute of the solution is solid silver nitrate, and the student will need to use a balance to measure the mass in grams. After adding the solid silver nitrate to the flask, the student will dissolve the solid silver nitrate in water, which is the solvent, and add enough water to the flask to reach a final volume of 100 milliliters of solution. We are told that the final concentration of the silver nitrate solution must be 0.1 molar. Our job in solving this problem is to determine the amount of silver nitrate in grams that the student must dissolve in water to make the 0.1-molar solution of silver nitrate.

The molar concentration, or molarity, of a solution is a measure of the number of moles of solute per liter of solution. In this problem, the units of molarity are written as a capital M. However, the units moles per liter are also often written to represent molarity.

During our problem-solving process, we’ll need to make use of the molarity equation. In this equation, 𝑛 is the number of moles of solute, 𝑐 is the molarity of the solution in moles per liter, and 𝑣 is the volume of the solution in liters. Before we can use the molarity equation, we need to convert the 100 milliliters of solution to liters. Then, we can substitute the molar concentration and volume of the solution into the equation to determine the amount of moles of silver nitrate. Our last step will be to convert the amount of moles of AgNO3 to grams.

Let’s begin by converting the volume of the solution from milliliters to liters. There are 1000 milliliters per one liter. To perform the conversion, we will need to multiply the volume of the solution, which is 100 milliliters, by one liter per 1000 milliliters. The result is 0.1 liters.

Now, we’re ready to substitute the concentration of the solution and the volume of the solution into the molarity equation. We can write 𝑛 equals 0.1 moles per liter multiplied by 0.1 liters. The result is 0.01 moles, which is the amount of moles of AgNO3. Because the amount in moles of a substance cannot be directly measured using a laboratory balance, we now need to convert the amount of moles of silver nitrate to grams.

We can make use of the following equation to help us. In this equation, 𝑛 is the number of moles, lowercase 𝑚 is the mass in grams, and capital 𝑀 is the molar mass in grams per mole. We will need to calculate the molar mass of silver nitrate first before we can use the equation.

The molar mass of silver nitrate can be calculated by adding together the average molar masses of the atoms that make up a formula unit of silver nitrate. One formula unit of silver nitrate is composed of one silver atom, one nitrogen atom, and three oxygen atoms. Next, we need to multiply the number of each type of atom by the average molar masses of the atoms, which are provided in the problem. After adding together the results of these calculations, we get an answer of 170 grams per mole, which is the molar mass of silver nitrate.

Now, we’re ready to substitute the amount of moles of silver nitrate, which was our answer from step two, and the molar mass of silver nitrate into the equation. We can write 0.01 moles equals lowercase 𝑚 divided by 170 grams per mole. We can rearrange the equation to solve for 𝑚. We get 𝑚 equals 170 grams per mole multiplied by 0.01 moles. After multiplying, the result is 1.7 grams. The student will need to dissolve 1.7 grams of silver nitrate to produce 100 milliliters of a 0.1-molar solution of silver nitrate.

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