How many moles of \(\mathrm{NaOH}\) are contained in \(27 \mathrm{~mL}\) of \(0.15 \mathrm{M} \mathrm{NaOH}\) ?

Short Answer

Expert verified
There are 0.00405 moles of \(\mathrm{NaOH}\) in 27 mL of 0.15 M \(\mathrm{NaOH}\).

Step by step solution

01

Understand the concept of molarity

Molarity (M) is a measure of the concentration of a solute in a solution, and it is defined as the number of moles of solute divided by the volume of solution in liters. The molarity equation is given by: \( M = \frac{n}{V} \), where \( M \) is the molarity, \( n \) is the number of moles of solute, and \( V \) is the volume in liters.
02

Convert the volume of NaOH solution to liters

Since the volume in the molarity formula needs to be in liters, convert the given volume from milliliters to liters. To convert from milliliters to liters, divide the volume by 1000. Thus, \( 27 \text{ mL} = \frac{27}{1000} \text{ L} = 0.027 \text{ L} \).
03

Calculate the number of moles of NaOH

Use the molarity equation to solve for \( n \): \( n = M \times V \). Plug the values of the molarity (0.15 M) and the volume in liters (0.027 L) into the equation to find the number of moles of NaOH: \( n = 0.15 \times 0.027 = 0.00405 \text{ moles} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Concentration of Solutions
Understanding the concentration of solutions is crucial in chemistry, as it allows us to describe how much solute is present in a certain amount of solvent. Molarity, often expressed with the symbol 'M', is one of the most common units used to express concentration. It is defined as the number of moles of solute divided by the volume of solution in liters.

The formula for molarity can be represented as:
\[ M = \frac{n}{V} \]
where \( M \) is molarity, \( n \) is the number of moles of the solute, and \( V \) is the volume of the solution in liters. When dealing with molarity, remember that the solution refers to the final mixture of solute plus solvent. In practical terms, when you have a 1 M solution of NaOH, it means that in every liter of the solution, there is 1 mole of sodium hydroxide present.

For exercises involving molarity, it's important to pay attention to the units given and ensure they are consistent with the units required for calculations. Inconsistent units can lead to incorrect results. Simple yet careful unit conversions are regularly the key to correct calculations in chemistry.
Moles Calculation
The mole is a fundamental unit in chemistry that provides a bridge between the atomic or molecular scale and the macroscopic scale. Calculating moles is essential for quantifying the amount of substances involved in chemical reactions.

The concept of the mole allows us to express quantities of atoms, ions, or molecules in manageable units, as counting individual particles is impractical. One mole is Avogadro's number (approximately \(6.022 \times 10^{23}\)) of particles, and it corresponds to the molecular weight of a substance expressed in grams.

To calculate moles from molarity, we use the following formula:
\[ n = M \times V \]
This equation reveals that the number of moles (\( n \)) is the product of molarity (\( M \)) and volume in liters (\( V \)). In the context of the exercise, we use the given molarity of the NaOH solution and the volume converted to liters to find the number of moles.

It's critical for students to understand that the mole concept is a way of translating between the mass of a substance and the number of particles contained in that mass, which becomes especially useful in stoichiometry and concentration calculations in chemistry.
Volume Conversion
Volume conversion is a necessary skill in chemistry as measurements may be provided in a variety of units, requiring conversion to a standard unit for accurate calculations. Commonly, volumes are measured in milliliters (mL) in laboratory settings, but many chemical equations and formulas, such as those for calculating molarity, require volume in liters (L).

To convert from milliliters to liters, remember that:\
\[ 1\,L = 1000\,mL \]
Therefore, to convert milliliters to liters, divide the number of milliliters by 1000. Conversely, to convert from liters to milliliters, you would multiply by 1000. For example, our exercise required converting 27 mL of NaOH solution to liters, resulting in:\
\[ 27\,\text{mL} = \frac{27}{1000}\,\text{L} = 0.027\,\text{L} \]
Grasping volume conversion is crucial because failing to convert volumes correctly can drastically affect the outcome of concentration calculations, leading to errors in molarity and the overall analysis of a chemical solution. Always double-check your conversion factor and the direction of your conversion (mL to L or L to mL) before applying it.

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Most popular questions from this chapter

In a reaction vessel \(0.184 \mathrm{~g}\) of \(\mathrm{NaOH}\) is required to be added for completing the reaction. How many millilitre of \(0.150 \mathrm{M} \mathrm{NaOH}\) solution should be added for this requirement.

(a) When \(4.2 \mathrm{~g} \mathrm{NaHCO}_{3}\) is added to a solution \(\mathrm{of} \mathrm{CH}_{3} \mathrm{COOH}\) weighing \(10.0 \mathrm{~g}\), it is observed that \(2.2 \mathrm{~g} \mathrm{CO}_{2}\) is released into atmosphere. The residue is found to weigh \(12.0 \mathrm{~g}\). Show that these observations are in agreement with the low of conservation of weigh. (b) If \(6.3 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3}\) are added to \(15.0 \mathrm{~g} \mathrm{CH}_{3} \mathrm{COOH}\) solution. The residue is found to weigh \(18.0 \mathrm{~g}\). What is the mass of \(\mathrm{CO}_{2}\) released in this reaction.

\(1.7 \mathrm{~g}\) of silver nitrate dissolved in \(100 \mathrm{~g}\) of water is taken. \(0.585 \mathrm{~g}\) of sodium chloride dissolved in \(100 \mathrm{~g}\) of water is added to it and chemical reaction occurs. \(1.435 \mathrm{~g}\) of silver chloride and \(0.85 \mathrm{~g}\) of sodium nitrate are formed. Justify that the data obey law of conservation of mass.

\(5.325 \mathrm{~g}\) sample of methyl benzoate, a compound used in the manufacture of perfumes is found to contain \(3.758 \mathrm{~g}\) of carbon, \(0.316\) \(\mathrm{g}\) hydrogen and \(1.251 \mathrm{~g}\) of oxygen. What is empirical formulia of compound. If mol. weight of methyl benzoate is \(136.0\), calculate its molecular formula.

Naturally occurring chlorine is \(75.53 \% \mathrm{Cl}^{35}\) which has an atomie mass of \(34.969 \mathrm{amu}\) and \(24.47 \% \mathrm{Cl}^{37}\), which has a mass of \(36.966\) amu. Calculate the average atomic mass of chlorine.

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