Find the molarity and molality of 37.0 wt% HCl. The density of a substance is the mass per unit volume. The table inside the back cover of this book tells us that the density of the reagent is 1.19 g/mL.
Normal alkanes are hydrocarbons with the formula CnH2n+2. Plants selectively synthesize alkanes with an odd number of carbon atoms. The concentration of C29H60 in summer rainwater collected in Hannover, Germany, is 34 ppb. Find the molarity of C29H60 and express the answer with a prefi x from Table 1-3.
The Correct Answer and Explanation is :
Let’s go step by step and solve these problems.
Problem 1: Molarity and Molality of 37.0 wt% HCl
- Given:
- Mass percentage of HCl = 37.0%
- Density of solution = 1.19 g/mL
- Molar mass of HCl = 36.46 g/mol
Step 1: Find Molarity (M)
Molarity (M) is defined as:
[
M = \frac{\text{moles of solute}}{\text{liters of solution}}
]
- Assume 1 L of solution:
- Mass of solution = ( 1.19 \times 1000 = 1190 ) g
- Mass of HCl = ( 0.37 \times 1190 = 440.3 ) g
- Moles of HCl = ( \frac{440.3}{36.46} = 12.08 ) moles
- Volume of solution = 1 L
[
M = \frac{12.08}{1} = 12.08 \text{ M}
]
Step 2: Find Molality (m)
Molality (m) is defined as:
[
m = \frac{\text{moles of solute}}{\text{kg of solvent}}
]
- Mass of solvent (water) = ( 1190 – 440.3 = 749.7 ) g = 0.7497 kg
- Molality:
[
m = \frac{12.08}{0.7497} = 16.1 \text{ m}
]
Thus, Molarity = 12.08 M and Molality = 16.1 m.
Problem 2: Molarity of C29H60 in Rainwater
- Given:
- Concentration = 34 ppb (parts per billion)
- Molar mass of C29H60 = 408.8 g/mol
- Assume density of water = 1 g/mL
Step 1: Convert ppb to g/L
[
34 \text{ ppb} = 34 \times 10^{-9} \text{ g per g of solution}
]
Since the density of water is 1 g/mL:
[
34 \times 10^{-9} \text{ g/mL} = 34 \times 10^{-6} \text{ g/L}
]
Step 2: Find Molarity
[
M = \frac{\text{mass of solute} (g/L)}{\text{molar mass} (g/mol)}
]
[
M = \frac{34 \times 10^{-6}}{408.8} = 8.32 \times 10^{-8} \text{ M}
]
Using prefixes, this is:
[
83.2 \text{ nM} \quad (\text{nano-molar})
]
Thus, the molarity of C29H60 in rainwater is 83.2 nM.
Explanation (300 words)
Understanding the concentration of chemical solutions is essential in both laboratory and environmental chemistry. Molarity (M) and molality (m) are two key concentration units. Molarity is the number of moles of solute per liter of solution, while molality is the number of moles per kilogram of solvent. In the case of 37.0 wt% HCl, the given density allows us to determine how much mass corresponds to 1 L of solution. Using these values, we calculate molarity as 12.08 M and molality as 16.1 m.
For the second problem, environmental chemists often measure organic compounds like alkanes in rainwater. The presence of hydrocarbons in rain is mainly due to atmospheric deposition from sources like plant waxes and pollution. The unit ppb (parts per billion) represents the ratio of the solute mass to the total mass of the solution. Since water density is approximately 1 g/mL, we convert 34 ppb of C29H60 into grams per liter and use the molar mass to determine its molarity, yielding 83.2 nM.
Both problems illustrate different approaches to handling concentration units. HCl, a strong acid, is present at a high concentration, while the hydrocarbon in rainwater exists in trace amounts. Such calculations are crucial in fields ranging from chemical manufacturing to environmental monitoring, where precise measurements impact industrial applications and ecological assessments.
Now, I’ll generate an image representing these concepts.
Here is the chemistry-themed infographic illustrating the concepts of molarity and molality for the 37.0 wt% HCl solution and the hydrocarbon concentration in rainwater. Let me know if you need any modifications or further explanations!
