Solving this problem requires us to use the formula for molarity, M = moles solute/liters of solution. We are given the molarity (0.1 M), the volume (750 ml), and need to calculate the number of moles and mass of NaOH required. Let’s break this down step-by-step:
Step 1: Convert Volume to Liters
The volume is provided in milliliters (ml), but molarity requires liters (L). So we first need to convert 750 ml to liters:
750 ml x (1 L/1000 ml) = 0.75 L
Step 2: Use Molarity Formula to Calculate Moles of NaOH
Now we can plug the values into the molarity formula:
M = moles solute/liters of solution
0.1 M = moles NaOH/0.75 L
To solve for moles NaOH, we rearrange the formula:
moles NaOH = M x liters of solution
moles NaOH = 0.1 M x 0.75 L
moles NaOH = 0.075 moles
Step 3: Use Moles to Calculate Grams of NaOH
Now that we know the number of moles of NaOH required, we can use the molar mass of NaOH to convert to grams.
The molar mass of NaOH is 40 g/mol.
Using the formula: grams = moles x molar mass
grams NaOH = 0.075 moles NaOH x 40 g/mol
grams NaOH = 3 g NaOH
Conclusion
To summarize, in order to make 750 ml of a 0.1 M NaOH solution, we need 3 grams of NaOH.
The key steps were:
- Convert volume to liters
- Use molarity formula to calculate moles of NaOH
- Use moles and molar mass to calculate grams of NaOH
Understanding these concepts of molarity and mole calculations is essential for chemistry students and professionals. Proper calculation of solution concentrations and reagents is critical for achieving accurate and repeatable results in the chemistry lab. With some practice, these calculations become second nature.
Sample Molarity Calculations
Let’s practice these calculations with a few more examples:
Example 1
How many grams of nitric acid (HNO3) are required to make 250 mL of a 0.5 M solution?
Step 1: Convert volume to liters
250 mL x (1 L/1000 mL) = 0.25 L
Step 2: Use molarity formula
M = moles/L
0.5 M = moles HNO3/0.25 L
Moles HNO3 = 0.125 moles
Step 3: Use moles and molar mass (63 g/mol) to calculate grams
Grams HNO3 = 0.125 moles x 63 g/mol = 7.9 g HNO3
Example 2
What volume of 2 M HCl solution can be made with 25 g HCl?
Step 1: Convert grams to moles
25 g HCl x (1 mol/36.5 g) = 0.685 moles HCl
Step 2: Rearrange molarity formula to solve for volume
M = moles/L
L = moles/M
L = 0.685 moles HCl/2 M = 0.34 L
So with 25 g HCl, we can make 0.34 L of 2 M HCl solution.
Example 3
If I have 325 mL of 0.25 M NaOH solution, how many moles of NaOH does it contain?
Step 1: Convert volume to liters
325 mL x (1 L/1000 mL) = 0.325 L
Step 2: Use molarity formula
M = moles/L
0.25 M = moles NaOH/0.325 L
Moles NaOH = 0.25 M x 0.325 L = 0.08125 moles NaOH
So there are 0.08125 moles of NaOH in the 325 mL of 0.25 M solution.
Practice Problems
Let’s apply these concepts to some practice problems. Try calculating the answers on your own before looking at the solutions.
Problem 1
How many grams of potassium chloride (KCl) are required to make 200 mL of a 0.1 M solution?
Solution:
Step 1: 200 mL x (1 L/1000 mL) = 0.2 L
Step 2: 0.1 M = moles KCl/0.2 L
moles KCl = 0.02
Step 3: Moles KCl x molar mass (74.5 g/mol) = 1.49 g KCl
Problem 2
What volume of 0.5 M NaOH can be made with 10 g of NaOH?
Solution:
Step 1: 10 g NaOH x (1 mol/40 g) = 0.25 moles NaOH
Step 2: M = moles/L
0.5 M = 0.25 moles NaOH/L
L = 0.25 moles/0.5 M = 0.5 L
So 10 g NaOH can produce 0.5 L of 0.5 M NaOH.
Problem 3
How many moles of solute are present in 250 mL of 0.1 M KCl solution?
Solution:
Step 1: 250 mL x (1 L/1000 mL) = 0.25 L
Step 2: 0.1 M = moles KCl/0.25 L
moles KCl = 0.025 moles
So there are 0.025 moles of KCl in the solution.
Tips for Solving Molarity Problems
When doing molarity calculations, keep these tips in mind:
- Convert all volumes to liters before using the molarity formula.
- Be sure you pay attention to the units and use the right molar mass when converting between mass and moles.
- Double check that your units cancel properly in each step.
- Use dimensional analysis to lay out the problem step-by-step and cancel units.
- Pay close attention to what quantity the problem is asking you to calculate. This will help you set up the correct molarity equation.
- Check your work! Plug your calculated values back into the original equation to ensure everything checks out.
Why is Molarity Important?
Now that we’ve gone over the basics of molarity calculations, let’s talk about why molarity is so important in chemistry.
Molarity is a conveniently scaled and standardized unit for concentration. By calculating and preparing solutions using molarity, chemists are able to reliably replicate experiments and processes. The molarity provides valuable information about the ratio of solute to solvent in the solution.
Some key reasons why molarity is important include:
- Describes solution composition: Molarity specifies the concentration of solute dissolved in a solution. This allows chemists to characterize the composition of a solution accurately.
- Allows replication of experiments: Solutions of specific molarities can be reliably reproduced, allowing consistent experimental conditions.
- Determines chemical properties: The molarity influences colligative properties such as boiling point elevation and freezing point depression. It impacts reaction kinetics and stoichiometry as well.
- Indicates reacting amounts: Using molarity, chemists can determine the amounts of reactants needed for reactions and stoichiometric calculations.
- Essential for titrations: Molarity is required to determine unknown concentrations in acid-base titrations and redox titrations.
In short, molarity is a cornerstone of solution chemistry. Mastering molarity calculations is essential for any chemistry student or professional.
Molarity in the Laboratory
To drive home the importance of molarity, let’s go over some examples of using molar solutions in a chemistry lab:
Preparing Reagents
Many laboratory procedures require reagents of specific molarities. For example, a common cell culture medium calls for 0.4 mM L-glutamine. By calculating the molarity correctly, a chemist can accurately prepare this reagent to support cell growth.
Making Standards
Analytical standards of known molar concentrations are essential for many instruments. For example, calibrating a UV-Vis spectrophotometer requires standard solutions of a compound at different molarities to generate a concentration curve.
Titration Analysis
Titration requires standardized molarity of titrant solutions. For instance, when titrating an unknown HCl solution with 0.1 M NaOH, the chemist must first accurately prepare the 0.1 M NaOH titrant.
Reaction Stoichiometry
To optimize chemical reactions, chemists use molarity to calculate the exact reagent ratios and amounts. For example, synthesizing a 100 g batch of biodiesel requires knowing the molar ratio of methanol to oil and calculating the volume of each at their given molarities.
Dilutions
Molarity allows determining the volume of stock needed to dilute to a target concentration. For example, 5 mL of 5 M HCl can be diluted to 100 mL to produce a 0.5 M HCl solution.
In all areas of the lab, from analytical chemistry to biochemistry, molarity is a fundamental concept underlying proper technique and calculation. Mastery of molarity is essential for any lab work involving solutions.
Molarity Calculations Quiz
Let’s test your understanding of molarity calculations with this brief quiz:
Quiz Question 1
How many moles of solute are present in 300 mL of 0.5 M NaCl solution?
A) 0.05 moles
B) 0.10 moles
C) 0.15 moles
D) 1.5 moles
Answer: B
Explanation: 0.5 M NaCl x 0.3 L = 0.10 moles NaCl
Quiz Question 2
What volume of 0.1 M HCl solution can be prepared with 25 g of HCl?
A) 0.25 L
B) 0.5 L
C) 1 L
D) 2.5 L
Answer: C
Explanation: 25 g HCl is 0.685 moles. 0.685 moles/0.1 M = 6.85 L
Quiz Question 3
How many grams of CaCl2 are needed to prepare 500 mL of 0.2 M solution?
A) 14.7 g
B) 29.4 g
C) 44.1 g
D) 73.5 g
Answer: B
Explanation: 0.5 L x 0.2 mol/L = 0.10 moles = 29.4 g CaCl2
How did you do? Being able to calculate molarity accurately takes practice, so review any concepts that you may be unclear on. Molarity calculations are a fundamental skill in chemistry.
Real World Applications of Molarity
We’ve mainly discussed molarity in an academic chemistry context. However, molarity has numerous real-world uses across many fields including:
Medicine
Doctors and pharmacists use molarity to calculate drug dosages and intravenous drip rates. Chemists develop analytical methods to verify drug purity and concentration.
Environmental Science
Environmental scientists measure pollutant concentrations in ppm and ppb, which are units related to molarity. They analyze water samples and calculate pollutant molarities.
Industry
Chemical engineers rely on molarity for process optimization and safety. Proper reagent molarities in large-scale reactions prevent accidents like runaway reactions.
Agriculture
Soil scientists use molarity to measure nutrient levels and acidity. Fertilizers and soil amendments are applied based on calculating molar concentrations.
Food Science
Food scientists standardize ingredients like salts, sweeteners, and acids using molar concentrations. This ensures consistent product quality and safety.
In many STEM careers, a thorough grasp of molarity and calculations is imperative. The principles discussed in this article broadly apply when working with solutions.
Molarity Problem Solving Tips
Here are some final tips for mastering molarity calculations:
- Memorize key molarity equations and unit conversions.
- Understand how to interconvert mass, moles, volume, and molarity.
- Practice doing word problems and setting up the equations.
- Check answers by plugging back into the original molarity relationship.
- Pay close attention to units and ensure they cancel properly.
- Use dimensional analysis and show work step-by-step.
- Ask for help if a concept or problem is unclear.
- Don’t get discouraged! Molarity takes practice to master.
With some time and effort, you can become adept at molarity calculations. It’s a foundational chemistry skill with many applications. Understanding molar concentrations will serve you well in your chemistry education and beyond.
