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Chapter 13: Problem 105

A lithium salt used in lubricating grease has the formula \(\mathrm{LiC}_{n} \mathrm{H}_{2 n+1} \mathrm{O}_{2} .\) The salt is soluble in water to the extent of 0.036 \(\mathrm{g}\) per 100 \(\mathrm{g}\) of water at \(25^{\circ} \mathrm{C}\) . The osmotic pressure of this solution is found to be 57.1 torr. Assuming that molality and molarity in such a dilute solution are the same and that the lithium salt is completely dissociated in the solution, determine an appropriate value of \(n\) in the formula for the salt.

Short Answer

Expert verified
The value of n in the formula for the lithium salt \(\mathrm{LiC}_{n}\mathrm{H}_{2n+1}\mathrm{O}_{2}\) is approximately 1, giving the molecular formula \(\mathrm{LiCH}_{3}\mathrm{O}_{2}\).

Step by step solution

01

Write down the osmotic pressure formula and given data

The osmotic pressure formula is given by: \(π = M * R * T\) where: π = osmotic pressure M = molarity R = gas constant (0.0821 L·atm/mol·K) T = temperature in Kelvin We are given: - solubility of lithium salt = 0.036 g per 100 g water at \(25^{\circ}\mathrm{C}\) - osmotic pressure (π) = 57.1 torr First, we need to convert the osmotic pressure from torr to atm and the temperature to Kelvin: - π = 57.1 torr × (1 atm / 760 torr) ≈ 0.0751 atm - T = \(25^{\circ}\mathrm{C} + 273\) = 298 K
02

Calculate the molality and molarity of the solution

We need to calculate the molality (which is assumed to be equal to molarity), given the solubility of lithium salt: molality = (mass of solute) / (1000 g solvent / molecular weight) Let's denote the molality as M and the molecular weight as MW, we can write: M = (0.036 g) / (1000 g solvent / MW)
03

Use osmotic pressure formula to determine molecular weight

Now, we can use the osmotic pressure formula to find the molecular weight of the lithium salt: π = M * R * T Substituting the values, 0.0751 atm = (0.036 g / (1000 g solvent / MW)) * 0.0821 L·atm/mol·K * 298 K Solving for MW, we get: MW ≈ 41.6 g/mol
04

Calculate the value of n using molecular weight

The molecular formula of the lithium salt is \(\mathrm{LiC}_{n}\mathrm{H}_{2n+1}\mathrm{O}_{2}\). Let's find the total molecular weight in terms of n: MW = 6.94 + 12.01n + 2.02(2n+1) + 2(16) Substitute the value of MW: 41.6 ≈ 6.94 + 12.01n + 4.04n + 4.04 + 32 Now, solve for n: n ≈ 1
05

Write down the final formula for the lithium salt

We found that the value of n equals 1. So, the molecular formula of the lithium salt is: \(\mathrm{LiC}_{1}\mathrm{H}_{2(1)+1}\mathrm{O}_{2}\) or \(\mathrm{LiCH}_{3}\mathrm{O}_{2}\)

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Most popular questions from this chapter

The partial pressure of \(\mathrm{O}_{2}\) in air at sea level is 0.21 atm. Using the data in Table \(13.1,\) together with Henry's law, calculate the molar concentration of \(\mathrm{O}_{2}\) in the surface water of a mountain lake saturated with air at \(20^{\circ} \mathrm{C}\) and an atmospheric pressure of 650 torr.

Indicate whether each statement is true or false: (a) A solute will dissolve in a solvent if solute-solute interactions are stronger than solute-solvent interactions. (b) In making a solution, the enthalpy of mixing is always a positive number. (c) An increase in entropy favors mixing.

Choose the best answer: A colloidal dispersion of one liquid in another is called (a) a gel, (b) an emulsion, (c) a foam, (d) an aerosol.

Lauryl alcohol is obtained from coconut oil and is used to make detergents. A solution of 5.00 g of lauryl alcohol in 0.100 kg of benzene freezes at \(4.1^{\circ} \mathrm{C} .\) What is the molar mass of lauryl alcohol from this data?

The maximum allowable concentration of lead in drinking water is 9.0 ppb. (a) Calculate the molarity of lead in a 9.0- ppb solution. (b) How many grams of lead are in a swimming pool containing 9.0 ppb lead in 60 \(\mathrm{m}^{3}\) of water?
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