Question

An experiment calls for \(1.00 \mathrm{~g}\) of silver nitrate, but all that is available in the laboratory is a \(0.50 \%\) solution of \(\mathrm{AgNO}_{3}\). Assuming the density of the silver nitrate solution to be very nearly that of water because it is so dilute, determine how many milliliters of the solution should be used.

Short answer

To obtain 1.00 g of silver nitrate from a 0.50% solution, 200 mL of the solution should be used.

Step-by-step solution

Convert percentage concentration to decimal fraction.

We are given that the concentration of the silver nitrate solution is 0.50%. To work with this more easily, we will convert the percentage to a decimal fraction by dividing by 100. So, \(0.50 \% = 0.50 / 100 = 0.005\) Now, we have the decimal fraction of the concentration as 0.005.

Determine the mass of silver nitrate in 1 mL of solution.

Since we are assuming the density of the silver nitrate solution to be very nearly that of water, we can assume that 1 mL of the solution weighs roughly 1 g. Therefore, we can find the mass of silver nitrate in 1 mL of the solution by multiplying its concentration (in decimal fractions) by the mass of 1 mL of the solution. Hence, Mass of silver nitrate in 1 mL of the solution = Concentration × Mass of 1 mL of solution \(= 0.005 \times 1\mathrm{~g}\) \(= 0.005\mathrm{~g}\) Now, we know that 1 mL of the solution contains 0.005 g of silver nitrate.

Calculate the required volume of the solution.

To obtain 1.00 g of silver nitrate, we can now find how many milliliters of the solution we need by dividing the required mass of silver nitrate (1.00 g) by the mass of silver nitrate in 1 mL of the solution (0.005 g). So, Required volume of the solution \(= \dfrac{\text{Required mass of silver nitrate}}{\text{Mass of silver nitrate in 1 mL of the solution}}\) \(= \dfrac{1.00\mathrm{~g}}{0.005\mathrm{~g/mL}}\) \(= 200\:\mathrm{mL}\) Therefore, 200 mL of the 0.50% silver nitrate solution should be used to obtain 1.00 g of silver nitrate.

Key concepts

Density

Density is a measure of how much mass is contained in a given volume. It is commonly expressed in units such as grams per cubic centimeter (g/cm³) or kilograms per liter (kg/L). When the exercise mentions that the density of the silver nitrate solution can be considered "very nearly that of water," it is implying that the solution is so dilute that its density is almost the same as pure water, which is approximately 1 g/cm³ or 1 kg/L. This assumption simplifies calculations by allowing us to directly equate the volume of liquid (in mL) to its mass (in g). For instance, when it says 1 mL of solution weighs roughly 1 g, this is adopting the assumption that the solution's density is close to that of water.

• Density determines how heavy a solution will be based on its volume.
• This is crucial in experiments where precision is needed, such as when measuring specific amounts of a chemical.

Understanding density helps us relate the mass of a solution to its volume, providing a seamless transition from concentration values to practical measurements in the lab.

Silver Nitrate

Silver nitrate, represented by the chemical formula \(\mathrm{AgNO}_3\), is an inorganic compound that is highly soluble in water. In laboratories, silver nitrate is often used in the solution form and has diverse applications both in chemistry experiments and industrial processes. Its importance in the exercise stems from the need to work with a specific concentration to achieve a desired reaction or result.

• Silver nitrate solutions are typically prepared in various concentrations depending on the requirement of the experiment.
• It is well-known for its photosensitive properties, turning black when exposed to light, which is a useful characteristic in photography and printing.
• In analytical chemistry, silver nitrate can be used for precipitating chloride ions in a solution as silver chloride.

Handling silver nitrate with care is vital due to its reactive nature, and accurate measurement with respect to its concentration influences the success of chemical processes or reactions.

Percentage Concentration

Percentage concentration expresses the amount of solute (in grams) per 100 mL of solution, offering a straightforward way to describe how concentrated a solution is. In the exercise, the 0.50% concentration indicates that there are 0.50 grams of silver nitrate in every 100 mL of solution. This form of concentration is particularly useful for quick calculations and conversions.

• To convert percentage concentration to a decimal fraction, divide by 100. For example, 0.50% becomes 0.005 when expressed as a decimal.
• This decimal can then be used to find the mass of solute per unit volume, allowing for practical measurement during experiments.
• Knowing the percentage concentration makes it easier to determine the volume of solution needed to obtain a certain mass of solute.

Mastery of percentage concentration allows you to calculate and prepare solutions accurately, essential in both laboratory experiments and industrial applications.