Naproxen is a commercially important anti-inflammatory agent that can be isolated from the thyroid gland. A solution of 1.138 g of naproxen in 25.00 g benzene has an osmotic pressure of 4.00 atm at 20°C. The density of benzene is 0.8787 g/mL at this temperature. Calculate the molar mass of naproxen, assuming it remains intact upon dissolution and the density of the solution equals the density of pure benzene. Naproxen is a commercially important anti-inflammatory agent that can be isolated from the thyroid gland. A solution of 1.138 g of naproxen in 25.00 g benzene has an osmotic pressure of 4.00 atm at 20°C. The density of benzene is 0.8787 g/mL at this temperature. Calculate the molar mass of naproxen, assuming it remains intact upon dissolution and the density of the solution equals the density of pure benzene. 230 g/mol 176 g/mol 307 g/mol 3.80 * 105 g/mol

Question

Chemistry

Lily West

20 September, 08:30

Naproxen is a commercially important anti-inflammatory agent that can be isolated from the thyroid gland. A solution of 1.138 g of naproxen in 25.00 g benzene has an osmotic pressure of 4.00 atm at 20°C. The density of benzene is 0.8787 g/mL at this temperature. Calculate the molar mass of naproxen, assuming it remains intact upon dissolution and the density of the solution equals the density of pure benzene. Naproxen is a commercially important anti-inflammatory agent that can be isolated from the thyroid gland. A solution of 1.138 g of naproxen in 25.00 g benzene has an osmotic pressure of 4.00 atm at 20°C. The density of benzene is 0.8787 g/mL at this temperature. Calculate the molar mass of naproxen, assuming it remains intact upon dissolution and the density of the solution equals the density of pure benzene. 230 g/mol 176 g/mol 307 g/mol 3.80 * 105 g/mol

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Answers (1)

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Niko Reilly · Answer 1

Molar mass for naproxen is 230 g/mol

Explanation:

Let's apply the colligative property to solve this:

π = M. R. T

4 atm = M. 0.082 l. atm/mol. K. 293K

4 atm / 0.082 mol. K/l. atm. 293K = M → 0.166 mol/L

Molarity are the moles of solute in 1L of solution.

Let's find out the mass of solution, in order to determine the volume (by density) and then, the moles we used.

Solvent + Solute = Solution → 1.138 g + 25 g = 26.138 g

Density of solution = Mass of solution/Volume of solution

Volume of solution = Mass of solute / Density of solution

26.138 g / 0.8787g/mL = 29.7 mL

We convert the volume from mL to L, to multiply by molarity

29.7 mL. 1L/1000 mL = 0.0297L → 0.166 mol/L. 0.0297L = 0.00494 moles

These are the moles, we used in the solution. Let's find out the molar mass → g/mol → 1.138 g / 0.00494 mol = 230.4 g/mol