Question

A sample of brine collected from the ocean has a density of \(1.28 \mathrm{~g} / \mathrm{mL}\) and contains \(7.55 \%\) sodium chloride. (a) What is the volume in \(\mathrm{L}\) of an \(8.14-\mathrm{kg}\) sample of this brine? (b) How many grams of brine are required to obtain \(150.0\) grams of sodium chloride?

Short answer

a) 6.359375 L. b) 1986.75 g brine.

Step-by-step solution

- Convert mass to volume for part (a)

Use the density formula: \ Density = \( \frac{mass}{volume} \) or \ \( \rho = \frac{m}{V} \). Rearrange to solve for volume: \ \( V = \frac{m}{\rho} \). The mass (m) is 8.14 kg, which converts to 8140 grams (since 1 kg = 1000 g). Use the density of brine \( \rho = 1.28 \, \text{g/mL} \). Volume in mL is: \( V = \frac{8140 \, \text{g}}{1.28 \, \text{g/mL}} = 6359.375 \, \text{mL} \). Convert to liters by dividing by 1000 (since 1 L = 1000 mL): \( V = \frac{6359.375 \, \text{mL}}{1000} = 6.359375 \, \text{L} \).

- Calculate mass of brine required for a given mass of NaCl for part (b)

First, find the percentage of NaCl in the brine: 7.55%. Note this means 7.55 grams of NaCl is present in every 100 grams of brine. Set up a proportion to find out how many grams of brine are needed to obtain 150.0 grams of NaCl: \ \( \frac{7.55 \, \text{g NaCl}}{100 \, \text{g brine}} = \frac{150.0 \, \text{g NaCl}}{x \, \text{g brine}} \). Cross-multiply and solve for x: \( 7.55x = 150.0 \times 100 \), \ \( 7.55x = 15000 \), \ \( x = \frac{15000}{7.55} = 1986.75496688 \, \text{g} \) or \( x \approx 1986.75 \, \text{g} \).

Key concepts

Density Calculations

Density is a fundamental concept in chemistry that relates the mass of an object to its volume. The formula for density is given by:
\[ \rho = \frac{m}{V} \]
where \( \rho \) is the density, \( m \) is the mass, and \( V \) is the volume. Understanding this relationship allows one to convert between mass and volume, which is crucial for solving many problems.
In the given exercise, the density of brine is 1.28 g/mL and the mass of the brine is 8.14 kg. To find the volume, we first convert the mass to grams, since the density is given in g/mL. Thus, 8.14 kg is 8140 grams (1 kg = 1000 grams).
Next, using the rearranged density formula, we get: \[ V = \frac{m}{\rho} = \frac{8140 \, \text{g}}{1.28 \, \text{g/mL}} = 6359.375 \, \text{mL} \]
To convert mL to L, we divide by 1000 (since 1 L = 1000 mL):
\[ V = \frac{6359.375 \, \text{mL}}{1000} = 6.359375 \, \text{L} \]
This result tells us that the volume of the brine sample is 6.359375 liters.

Percent Composition

Percent composition is the percentage by mass of each element or component in a compound or mixture. In this exercise, the brine solution contains 7.55% sodium chloride (NaCl). This means that in every 100 grams of brine, there are 7.55 grams of NaCl.
To solve part (b) of the exercise, we need to determine how many grams of brine are required to obtain 150.0 grams of NaCl. Using the percentage composition, we set up a proportion:
\[ \frac{7.55 \, \text{g NaCl}}{100 \, \text{g brine}} = \frac{150.0 \, \text{g NaCl}}{x \, \text{g brine}} \]
Cross-multiplying to solve for \( x \):
\[ 7.55x = 150.0 \times 100 \]
\[ 7.55x = 15000 \]
Finally, dividing both sides by 7.55:
\[ x = \frac{15000}{7.55} = 1986.75496688 \, \text{g} \]
Rounded to two decimal places, we get:
\[ x \approx 1986.75 \, \text{g} \]
Therefore, 1986.75 grams of brine are required to obtain 150.0 grams of sodium chloride.

Unit Conversion

Unit conversion is essential in chemistry to ensure that measurements are in the correct units for calculations. In our exercise, we had to convert kilograms to grams and milliliters to liters.
For mass, remember that:

• 1 kilogram (kg) = 1000 grams (g)

This means to convert kilograms to grams, you simply multiply by 1000.
For volume, understand that:

• 1 liter (L) = 1000 milliliters (mL)

Hence, to convert milliliters to liters, you divide by 1000.
These conversions were necessary to ensure that the units of density (1.28 g/mL) matched the units of mass and volume in our calculations. Making conversion errors can lead to incorrect results, so always double-check your units!
Mastering unit conversion helps in solving a wide range of chemistry problems accurately and efficiently.