How many grams of KBr are needed to prepare 650 ml of a 0.115 M KBr solution?

To determine the grams of KBr required to make a 0.115 M KBr solution with a volume of 650 ml, we need to use the formula:

Table of Contents

The Formula

Molarity (M) = moles of solute (n) / Volume of solution (L)

M = n/V

Using this formula, we can calculate the number of moles of KBr needed for our desired molarity and volume. Then, knowing the moles of KBr, we can convert to grams using the molar mass of KBr.

Calculating Moles of KBr

First, we’re given the desired molarity is 0.115 M KBr.

M = 0.115 moles/L

We’re also given the desired volume is 650 ml. We need to convert this to liters:

650 ml x (1 L / 1000 ml) = 0.650 L

Now we can plug these values into the formula to calculate the moles of KBr required:

M = n/V

0.115 moles/L = n / 0.650 L

n = (0.115 moles/L) x (0.650 L) = 0.0748 moles KBr

So, for a 0.115 M solution with a volume of 650 ml, we need 0.0748 moles of KBr.

Converting Moles to Grams

Now that we know the number of moles of KBr required, we can convert from moles to grams using the molar mass.

The molar mass of KBr is:

K: 39.1 g/mol

Br: 79.9 g/mol

So the molar mass of KBr is:

KBr: 39.1 g/mol + 79.9 g/mol = 119 g/mol

Using the molar mass, we can convert moles KBr to grams KBr:

0.0748 moles KBr x (119 g/mol) = 8.90 grams KBr

Summary

To prepare 650 ml of a 0.115 M KBr solution:

Calculate moles of KBr needed using the molarity formula:
- Molarity = 0.115 M
- Volume = 650 ml = 0.650 L
- Moles KBr = (0.115 mol/L) x (0.650 L) = 0.0748 mol
Convert moles of KBr to grams using molar mass:
- Molar mass KBr = 119 g/mol
- Grams KBr = (0.0748 mol) x (119 g/mol) = 8.90 g

Therefore, 8.90 grams of KBr are required to prepare 650 ml of a 0.115 M KBr solution.

Showing the Calculations Step-by-Step

Here are the calculations step-by-step:

Start with the given:
- Desired molarity = 0.115 M
- Desired volume = 650 ml
Convert volume to liters:
- 650 ml x (1 L / 1000 ml) = 0.650 L

Use molarity formula to calculate moles of KBr:
- M = n/V
- 0.115 mol/L = n / 0.650 L
- n = (0.115 mol/L) x (0.650 L)
- n = 0.0748 mol KBr
Calculate molar mass of KBr:
- K: 39.1 g/mol
- Br: 79.9 g/mol
- KBr: 39.1 g/mol + 79.9 g/mol = 119 g/mol

Convert moles of KBr to grams:
- Grams KBr = (0.0748 mol) x (119 g/mol)
- Grams KBr = 8.90 g

Following these step-by-step calculations confirms that 8.90 grams of KBr are required to make 650 ml of 0.115 M KBr solution.

Explaining Each Step in Detail

Here is a more detailed explanation of each step:

We are given the desired molarity of the KBr solution, which is 0.115 M. Molarity (M) is a concentration unit that measures the number of moles of solute (in this case KBr) per liter of solution. We’re also told the desired volume is 650 ml.

Since molarity uses liters, we need to convert the volume of 650 ml to liters. To do this, we use the unit conversion:
- 1000 ml = 1 L
So 650 ml x (1 L / 1000 ml) = 0.650 L

Now we can use the molarity formula to calculate the number of moles of KBr needed:
- Molarity (M) = moles of solute (n) / liters of solution (L)
- Plug in the values we know:
  - M = 0.115 mol/L
  - L = 0.650 L
- Rearrange and solve:
  - 0.115 mol/L = n / 0.650 L
  - n = (0.115 mol/L) x (0.650 L)
  - n = 0.0748 mol KBr

Now that we know the number of moles of KBr required, we need to convert moles to grams using the molar mass.
- First, we calculate the molar mass of KBr by adding the molar masses of potassium (K) and bromine (Br):
  - K: 39.1 g/mol
  - Br: 79.9 g/mol
  - KBr: 39.1 g/mol + 79.9 g/mol = 119 g/mol
- Then, use the molar mass to convert moles to grams:
  - Grams KBr = (0.0748 mol KBr) x (119 g/mol)
  - Grams KBr = 8.90 g

By methodically going through each step, we determine the grams of KBr required is 8.90 grams.

Why Molarity is Used

Molarity (M) is commonly used to prepare solutions because it provides an exact concentration in terms of moles per liter. Some key reasons why molarity is useful include:

Molarity provides a quantitative measure of solute concentration, rather than qualitative terms like “dilute” or “concentrated”.
The molar amount (moles) of solute is independent of the solute’s identity. This allows comparing solutions of different solutes by molarity.

A molar solution contains an easily convertible and measurable amount of solute (moles) per unit volume (liters).
Molarity automatically accounts for volume changes. If you dilute a 1 M solution by adding more solvent, the molarity decreases proportionally.
Many chemical reactions depend on molar amounts of reactants, so molarity provides the right stoichiometric proportions.

For lab experiments, manufacturing, and other chemistry applications, molarity provides a convenient way to make solutions with precise concentrations of solute. The molar amounts can be easily scaled up or down while maintaining the same molarity.

How to Make Solutions of Desired Molarity

Use the following general procedure to make a solution of desired molarity:

Calculate the target number of moles of solute needed using the molarity formula:

Molarity (M) = moles solute (n) / liters solution (L)
Rearrange to solve for n:
n = (Molarity) x (Liters of solution)

Weigh out the corresponding mass of solute based on its molar mass:

Use molar mass (g/mol) to convert moles solute to grams
Use a balance to measure the calculated grams of solute

Transfer the solute to a volumetric flask and add solvent up to the calibration line. This ensures the final volume matches your target.
Mix thoroughly to dissolve the solute and create a homogeneous solution.
Optional: Confirm molarity using analytical techniques like titration.

Following this procedure allows precise preparation of molar solutions for a variety of solutes and concentrations. The same approach applies whether making a laboratory-scale solution or industrial quantities.

How Molarity Changes with Dilution

An important point about molarity is that it changes when a solution is diluted. For example, if you have 1 liter of a 1 M NaCl solution, adding 1 more liter of water will change it to a 0.5 M NaCl solution. Here’s why:

Initially:
- Volume = 1 L
- Moles NaCl = 1 mol (based on 1 M)
After adding 1 L water:
- Volume = 2 L
- Moles NaCl is still 1 mol (doesn’t change)
New molarity:
- M = moles / volume
- M = (1 mol) / (2 L) = 0.5 M

So diluting by adding solvent decreases molarity, while evaporating solvent increases molarity. This relationship allows calculating new molarities when solutions are diluted.

Using Molarity in Calculations

Molarity (M) can be used in various stoichiometric calculations involving solutions. Here are some examples:

Calculating reactant/product amounts in reactions
E.g. How many grams of NaCl are produced from 0.75 L of 2.0 M NaOH reacting with 1.5 L of 1.0 M HCl?
Determining solution concentrations

Finding dilution factors to prepare new molarities
Determining solution volumes to contain given moles of solute
Comparing concentrations of different solutes

The key is converting between moles, volume, molarity, and mass using dimensional analysis and the molarity equation. This makes molarity a versatile tool for aqueous chemistry calculations.

Limitations of Molarity

While very useful, molarity does have some limitations:

Only applies to dilute solutions, where volume change on mixing is negligible.

Doesn’t account for intermolecular interactions that affect dissolution.
Assumes solution volume is exact, although pipetting errors can occur.
Doesn’t indicate degree of ionization of solutes.

Changes with temperature due to thermal expansion/contraction.

For very concentrated solutions, molality (mols solute per kg solvent) is preferred over molarity. And molarity cannot capture molecular-level deviations from ideality.

However, for most common laboratory and industrial applications, molarity provides an easy and standardized way to express solution concentrations. The approximations hold reasonably well for dilute solutes up to around 1 M concentration.

Conclusion

In summary, molarity is a convenient concentration unit for preparing solutions by providing the moles of solute per liter of solution. It can be easily used for stoichiometric calculations and conversions between moles, mass, and solution volumes. By understanding how to calculate molarity and how it changes with dilution, many solution-based calculations become straightforward. While it has some limitations, molarity provides a useful standard approach for working with solution concentrations in chemistry and lab experiments.