Question

Household ammonia used for cleaning contains about \(10 \mathrm{~g}\) (two significant figures) of \(\mathrm{NH}_{3}\) in \(100 \mathrm{~mL}\) (two significant figures) of solution. What is the molarity of the \(\mathrm{NH}_{3}\) in solution?

Short answer

Answer: The molarity of NH3 in the solution is approximately 5.87 M.

Step-by-step solution

Find moles of NH3

First, find the moles of NH3 using the mass and molar mass: Moles of NH3 = mass of NH3 / molar mass of NH3 The mass of NH3 is given as 10 g, and the molar mass of NH3 is 14.01 g/mol for nitrogen and 3 * 1.01 g/mol for three hydrogen atoms, which adds up to 17.03 g/mol. Moles of NH3 = (10 g) / (17.03 g/mol)

Calculate the volume of the solution in Liters

We are given that the volume of the solution is 100 mL. To convert this to Liters, we use the conversion factor: 1 L = 1000 mL. Volume in Liters = (100 mL) / (1000 mL/L) = 0.1 L

Calculate the molarity of NH3

Now that we have the moles of NH3 and the volume of the solution in Liters, we can find the molarity by dividing the moles by the volume. Molarity of NH3 = moles of NH3 / volume in Liters Molarity of NH3 = (10 g) / [(17.03 g/mol) * (0.1 L)]

Solve and report the molarity to the appropriate number of significant figures

When solving the equation above, make sure to consider significant figures. Since both the mass and the volume of the solution are given to two significant figures, the molarity of NH3 should also be reported to two significant figures. Molarity of NH3 ≈ 5.87 M (rounded to two significant figures)

Key concepts

Ammonia Solution

Ammonia is a compound composed of nitrogen and hydrogen atoms, with the chemical formula \( \mathrm{NH}_3 \). It is a colorless gas that is highly soluble in water, creating what is known as an ammonia solution. This solution is often used as a cleaning agent in households.
In the context of the exercise, ammonia solution refers to the mixture made by dissolving 10 grams of \( \mathrm{NH}_3 \) in 100 mL of water. The concentration of this solution can be expressed in terms of molarity. Understanding how to compute the molarity allows us to quantify the concentration of ammonia, which is crucial for both safety and efficacy in its application.

Moles and Molar Mass

To determine the molarity of ammonia in the solution, we first need to calculate the number of moles of \( \mathrm{NH}_3 \). The concept of moles is fundamental in chemistry, acting as a bridge between the mass of a substance and the number of particles it contains.
The molar mass is the weight of one mole of a compound and is specific to each substance. For \( \mathrm{NH}_3 \), the molar mass is calculated by adding the atomic masses of nitrogen and hydrogen:

• Nitrogen (\( \mathrm{N} \)) has a molar mass of 14.01 g/mol.
• Hydrogen (\( \mathrm{H} \)) has a molar mass of 1.01 g/mol. Since there are three hydrogen atoms, this adds to 3.03 g/mol.
• The total molar mass of \( \mathrm{NH}_3 \) is thus 17.03 g/mol.

To find the moles, we divide the mass of \( \mathrm{NH}_3 \) by its molar mass:

Significant Figures

Significant figures are essential when reporting scientific calculations, as they indicate the precision of measured quantities. In this exercise, both the mass of \( \mathrm{NH}_3 \) (10 g) and the volume of the solution (100 mL) are given to two significant figures, which means our final answer for molarity should respect this level of precision to reflect the certainty of the data.
Using significant figures correctly entails:

• Identifying all non-zero digits as significant.
• Including zeros that are between non-zero digits.
• Avoiding zeros used solely for spacing the decimal point (leading or trailing zeros in a decimal figure when not in the middle).

When performing calculations, the result should be rounded to the least number of significant figures present in the given data. So, the molarity should be reported as 5.87 M, rounded to two significant figures.

Volume Conversion

In many scientific calculations involving concentration, the volume must be expressed in liters (L). This is because molarity is defined as moles per liter. Initially, the problem states the volume of the ammonia solution as 100 mL.
To convert this to liters, we use the relationship:

• 1 liter (L) = 1000 milliliters (mL).

Thus, the conversion from milliliters to liters is straightforward:

• Divide the number of milliliters by 1000.
• In this exercise, 100 mL is equal to 0.1 L.

This conversion is crucial as it allows us to use the standard units of molarity and accurately calculate the final concentration of \( \mathrm{NH}_3 \) in the solution.