\(100 \mathrm{~g}\) of \(\mathrm{NaCl}\) is stirred in \(100 \mathrm{~mL}\) of water at \(20^{\circ} \mathrm{C}\) till the equilibrium is attained : (a) How much \(\mathrm{NaCl}\) goes into the solution and how much of it is left undissolved at equilibrium? The solubility of \(\mathrm{NaCl}\) at \(20^{\circ} \mathrm{C}\) is \(6.15\) mol/litre. (b) What will be the amount of \(\mathrm{NaCl}\) left undissolved if the solution is diluted to \(200 \mathrm{~mL} ?\)

Short Answer

Expert verified
At equilibrium, 36.90 g of NaCl will be dissolved, and 63.10 g will be left undissolved. After dilution to 200 mL, still 63.10 g of NaCl will be undissolved, as dilution doesn't affect the amount of saturated solute.

Step by step solution

01

Calculate moles of NaCl initially

Convert the given mass of NaCl to moles using its molar mass (58.44 g/mol). Moles of NaCl = mass (g) / molar mass (g/mol) = 100 g / 58.44 g/mol.
02

Determine moles of NaCl that can be dissolved

Use the solubility information to calculate the number of moles that can be dissolved in 100 mL of water. Since the solubility is given in mol/L, convert 100 mL to liters by dividing by 1000. Solubility in 100 mL = solubility (mol/L) * volume (L) = 6.15 mol/L * 0.1 L.
03

Calculate the moles of NaCl that will be undissolved at equilibrium

Subtract the moles that can be dissolved from the initial moles of NaCl to find the moles of undissolved NaCl at equilibrium. Undissolved NaCl (moles) = initial moles - moles dissolved.
04

Convert the undissolved moles back to mass

Multiply the moles of undissolved NaCl by the molar mass to get the mass of NaCl that is undissolved at equilibrium. Undissolved NaCl (g) = undissolved NaCl (moles) * molar mass (g/mol).
05

Calculate the new equilibrium after dilution

Diluting the solution to 200 mL will not change the amount of NaCl already dissolved (still limited by its solubility), so the moles of NaCl undissolved will remain the same. Thus, the mass of undissolved NaCl also remains unchanged after dilution.

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Key Concepts

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

Molar Mass Calculation
Understanding how to calculate the molar mass is a fundamental skill in chemistry. The molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule.

For instance, sodium chloride (NaCl) consists of sodium (Na) and chlorine (Cl) ions. Using the periodic table, the atomic mass of Na is approximately 22.99 g/mol, and that of Cl is about 35.45 g/mol. The molar mass of NaCl, then, is the sum of the atomic masses of Na and Cl, or 58.44 g/mol.

To elucidate the process with an example: if we have 100 g of NaCl and want to calculate the number of moles, we use the formula:
\[\begin{equation}\text{number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{100 \text{ g}}{58.44 \text{ g/mol}}.\end{equation}\]

This step is critical as it bridges the gap between mass and the number of molecules, which is essential in comprehending chemical reactions and solution concentrations.
Moles to Mass Conversion
Once the number of moles in a substance is known, the conversion to mass is just a mathematical step away. The molar mass serves as a conversion factor between moles and grams.

To convert moles to mass, multiply the number of moles by the molar mass of the substance, as shown in this equation:
\[\begin{equation}\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}.\end{equation}\]

In the context of our example with NaCl, if we have leftover NaCl in undissolved form at equilibrium and we know the moles of it, we can determine the mass that remains by using NaCl's molar mass. This step is fundamental for quantifying substances in the laboratory and can be applied to various applications, including the analysis of reaction yields and the preparation of solutions with precise concentrations.
Solution Dilution Effects
Dilution refers to the process of reducing the concentration of a solute in a solution, usually by adding more solvent. This is frequently done in chemistry to achieve desired concentrations for various experimental needs. However, it's important to note that dilution doesn't change the quantity of solute present; it simply spreads it out in a larger volume of solvent.

In the case of our NaCl solution, when we dilute it from 100 mL to 200 mL, the total amount of dissolved NaCl does not change because the solubility limit remains constant. Solubility is a characteristic property of a substance and defines how much of it can dissolve in a given amount of solvent at a specific temperature.

Here's a critical point to grasp:
  • Before dilution, we calculate the mass of undissolved NaCl at equilibrium using its solubility in 100 mL of water.
  • After dilution, since the solubility remains the same, the undissolved amount will not change. The previously dissolved NaCl is now simply present in a larger volume of water.

Understanding these effects is crucial when preparing solutions for chemical reactions or titrations, where the concentration directly influences the outcome of the experiment.

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Most popular questions from this chapter

Write euilibrium constant for the each : (a) \(\quad \mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(\mathrm{~g})}\) (b) \(\quad \mathrm{KClO}_{3(\mathrm{~s})} \rightleftharpoons \mathrm{KCl}_{(\mathrm{s})}+(3 / 2) \mathrm{O}_{2(\mathrm{~g})}\) (c) \(\mathrm{CaC}_{2(\mathrm{~s})}+5 \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{CaCO}_{3(\mathrm{~s})}+2 \mathrm{CO}_{2(\mathrm{~g})}\) (d) \(\mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}\) (e)Fe \(^{3+}{ }_{\text {(aq })}+\mathrm{SCN}_{\text {(aq.) }}^{*}=\mathrm{Fe}(\mathrm{SCN})^{2+}{ }_{\text {(aq })}\) (f) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})} \rightleftharpoons \mathrm{CuSO}_{4(\mathrm{~s})}+5 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{v})}\)

For the reaction; \(\mathrm{N}_{2 \mathrm{k} \mathrm{g}}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})}\) At \(400 \mathrm{~K}, K_{\mathrm{p}}=41 \mathrm{~atm}^{-2}\). Find the value of \(K_{\mathrm{p}}\) for each of the following reactions at the same temperature: (i) \(2 \mathrm{NH}_{3(\mathrm{~g})} \Longrightarrow \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})}\) (ii) \({ }_{2} \mathrm{~N}_{2(\mathrm{~g})}+\frac{3}{2} \mathrm{H}_{2(g)} \Longrightarrow \mathrm{NH}_{3(\mathrm{~g})}\); (iii) \(2 \mathrm{~N}_{2(\mathrm{~g})}+6 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 4 \mathrm{NH}_{3(\mathrm{~g})}\)

Sodium metal crystallizes in a body centred cubic lattice with the cell edge \(a=4.29 \AA\). What is the radius of sodium atom?

If a mixture of 3 moles of \(\mathrm{H}_{2}\) and one mole of \(\mathrm{N}_{2}\) is completely converted into \(\mathrm{NH}_{3}\). What would be the ratio of the initial and final volume at same temperature and pressure?

Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below: $$ 2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NOBr}_{(\mathrm{g})} $$ When \(0.087\) mole of \(\mathrm{NO}\) and \(0.0437\) mole of \(\mathrm{Br}_{2}\) are mixed is a closed container at constant temperature, \(0.0518\) mole of NOBr is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine.

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