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Chapter 2: Problem 92

\(4 \mathrm{~g}\) of \(\mathrm{NaOH}\) are present in \(0.1 \mathrm{dm}^{3}\) solution have specific gravity \(1.038 \mathrm{~g} / \mathrm{mL}\). Calculate : (a) mole fraction of \(\mathrm{NaOH}\); (b) molality of \(\mathrm{NaOH}\) solution; (c) molarity of \(\mathrm{NaOH}\) solution; (d) normality of \(\mathrm{NaOH}\) solution.

Short Answer

Expert verified
The mole fraction of NaOH is approximately 0.0177, the molality is approximately 1.003 mol/kg, the molarity is 1 M, and the normality is also 1 N.

Step by step solution

01

Calculate Mole Fraction of NaOH

To calculate the mole fraction of NaOH, we need to know the moles of NaOH and the total moles of the solution. First, calculate the moles of NaOH using its molar mass, which is approximately 40 g/mol. Moles of NaOH = mass (g) / molar mass (g/mol) = 4 g / 40 g/mol = 0.1 moles. Next, calculate the mass of the solution using its volume and specific gravity. Mass of solution = volume (dm^3) * specific gravity (g/mL) * 1000 (to convert dm^3 to mL) = 0.1 dm^3 * 1.038 g/mL * 1000 = 103.8 g. Now, convert the mass of solution to moles by subtracting the moles of NaOH and dividing by the molar mass of water (approximately 18 g/mol). Moles of water = (mass of solution - mass of NaOH) / molar mass of water = (103.8 g - 4 g) / 18 g/mol ≈ 5.544 moles. Finally, calculate the mole fraction of NaOH = moles of NaOH / (moles of NaOH + moles of water) = 0.1 / (0.1 + 5.544).
02

Calculate Molality of NaOH Solution

Molality is calculated by the number of moles of the solute per kilogram of the solvent. We have already calculated the moles of NaOH as 0.1 moles. Mass of solvent (water) in kilograms is the total mass of the solution minus the mass of NaOH. Mass of solvent (kg) = (103.8 g - 4 g) / 1000 = 0.0998 kg. Therefore, molality (m) = moles of solute / mass of solvent (kg) = 0.1 moles / 0.0998 kg.
03

Calculate Molarity of NaOH Solution

Molarity is defined as moles of solute per liter of solution. We have already found the number of moles of NaOH to be 0.1 moles. The volume of the solution in liters is 0.1 dm^3, which is equal to 0.1 liters. Thus, molarity (M) = moles of NaOH / volume of solution (L) = 0.1 moles / 0.1 L.
04

Calculate Normality of NaOH Solution

Normality (N) is the moles of equivalent solute per liter of solution. For NaOH, which is a monoprotic base, the equivalent weight equals its molecular weight. Therefore, in this case, the normality of the NaOH solution is the same as its molarity. Hence, normality (N) = molarity (M) = 0.1 moles / 0.1 L.

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Key Concepts

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

Mole Fraction
The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the number of moles of a particular constituent to the total number of moles of all constituents in the mixture. Let's say we're dealing with a NaOH solution with water as the solvent. To find the mole fraction of NaOH, we divide the moles of NaOH by the sum of the moles of NaOH and the moles of water.

To illustrate:
  • Determine the number of moles of NaOH.
  • Calculate the number of moles of water by accounting for its mass and molar mass.
  • Add the moles of NaOH and moles of water to find the total moles in the solution.
  • The mole fraction of NaOH is then the moles of NaOH divided by the total moles.
This value is dimensionless and is useful in various calculations, particularly those involving the physical properties of solutions, like boiling point elevation or freezing point depression.
Molality
Moving on to molality, this is a concentration term referring to the moles of solute per kilogram of solvent. Unlike molarity, molality is not affected by temperature changes because it's based on the mass of the solvent, not the volume of the solution. For a NaOH solution, calculate molality with the following steps:

  • From the mass of NaOH, calculate the moles of NaOH.
  • Find the mass of the solvent (water) in kilograms by subtracting the mass of NaOH from the total mass of the solution.
  • Divide the moles of NaOH by the mass of the solvent in kilograms to get molality.
Because molality is based on mass, it's particularly useful in situations where temperature and pressure may vary.
Molarity
The term molarity is a commonly used measure of concentration in chemistry. It is defined as the moles of solute per liter of solution. To find the molarity of the NaOH solution:
  • Capture the number of moles of NaOH calculated previously.
  • Ascertain the volume of the solution in liters (often given in the problem or easy to convert).
  • Divide the moles of NaOH by the volume of the solution in liters for the molarity.
Molarity is extensively used in stoichiometry and reacting quantities calculations as it directly relates the volume of solution to the number of moles of a solute.
Normality
Lastly, we have normality, which is similar to molarity but takes into account the equivalent factor of the solute. It is the number of gram equivalents of solute per liter of solution. To find the normality of a NaOH solution:
  • First, note that for NaOH, a monoprotic base, the equivalent weight is the same as its molecular weight.
  • Since we have the number of moles and NaOH is monoprotic, the normality equals the molarity in this specific case.
  • Divide the moles of NaOH by the volume of the solution in liters to get the normality, which is the same as molarity here.
Normality is particularly useful in acid-base titrations and precipitation reactions where ions combine in simple ratios.

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Most popular questions from this chapter

How would you prepare exactly \(3.0\) litre of \(1.0 \mathrm{M} \mathrm{NaOH}\) by mixing proportions of stock solutions of \(2.50 \mathrm{M} \mathrm{NaOH}\) and \(0.40 \mathrm{M} \mathrm{NaOH}\). No water is to be used.

Potassium chromate is isomorphous to potassium sulphate \(\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right)\) and it is found to have \(26.78 \%\) Cr. Calculate the at.wt. of \(\mathrm{Cr}\) if at.wt. of potassium is \(39.10\).

A mixture of \(1.0\) mole of \(\mathrm{Al}\) and \(3.0\) mole of \(\mathrm{Cl}_{2}\) are allowed to react as: $$ 2 \mathrm{Al}_{(\mathrm{s})}+3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{AlCl}_{3(\mathrm{~s})} $$ (a) Which is limiting reagent? (b) How many moles of \(\mathrm{AlCl}_{3}\) are formed? (c) Moles of excess reagent left unreacted.

What are the final concentrations of all the ions when following are mixed? \(50 \mathrm{~mL}\) of \(0.12 \mathrm{M} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}, 100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{FeCl}_{3}\) and \(100 \mathrm{~mL}\) of \(0.26 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\)

Carbon and oxygen are known to form two compounds. The carbon content in one of these is \(42.9 \%\) while in the other it is \(27.3 \%\). Show that this data is in agreement with the law of multiple proportions.
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