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.

To find the value of 'n', we will use the osmotic pressure formula, which is given by: π = MRT where π is the osmotic pressure, M is the molarity, R is the gas constant (0.0821 atm·L/mol·K), and T is the temperature in Kelvin (25°C = 298.15 K).

Since the solubility of the salt is given as 0.036 g per 100 g of water, and considering the density of water to be 1 g/ml, we can write the mass of the salt per liter of solution and convert it into moles per liter (molarity) using the formula weight of the salt. The mass of salt in 1 L of solution = (0.036 g / 100 g of water) × (1000 g of water / 1 L of solution) = 0.36 g of salt in 1 L of solution. The formula weight of the lithium salt = Li + Cₙ + H₂ₙ+₁ + 2O The molar mass of Li = 6.94 g/mol, C = 12.01 g/mol, H = 1.01 g/mol, and O = 16.00 g/mol The formula weight of the lithium salt = 6.94 + 12.01n + (2ₙ+₁) × 1.01 + 2 × 16.00 Now we can calculate molarity (M) as: M = (0.36 g of salt in 1 L of solution) / (6.94 + 12.01n + 2.02(ₙ+1) + 32.00 g/mol)

Since the osmotic pressure is given in torr, we need to first convert it to atmospheres (atm), as it is the relevant unit for the osmotic pressure formula. Osmotic pressure in atmospheres = 57.1 torr × (1 atm / 760 torr) ≈ 0.075 atm

Now, we can plug the osmotic pressure, molarity, and temperature into the osmotic pressure formula, and solve for n. π = MRT 0.075 atm = (0.36 g of salt in 1 L of solution) / (6.94 + 12.01n + 2.02(ₙ+1) + 32.00 g/mol) × (0.0821 atm·L/mol·K) × 298.15 K Dividing both sides by 0.0821 and 298.15, we get: 0.00304 mol/L = (0.36 g of salt in 1 L of solution) / (6.94 + 12.01n + 2.02(ₙ+1) + 32.00 g/mol) Solving for 'n', we get approximately: n ≈ 3.0

Since 'n' is an integer value, we can round it to the nearest integer, which is 3. Thus, the appropriate value of n in the formula for the lithium salt is 3. The formula for the lithium salt would be LiC₃H₇O₂.