A solution contains 12.00 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume...

A solution contains 12.00 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -4.95 ∘C. The mass percent composition of the compound is 53.31% C, 11.19% H, and the rest is O. Part A What is the molecular formula of the compound? Express your answer as a molecular formula.

Expert Solution

queen_honey_blossom answered 3 months ago

A solution contains 10.75g of an unknown compound dissolved in 50.0 mL of water. (Assume a...

A solution contains 10.75g of an unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.38 C. The mass percent composition of the compound is 60.98% C, 11.94% H, and the rest is O. -What is the molecular formula of the compound? Express your answer as a molecular formula..

A solution contains 10.10 gg of unknown compound dissolved in 50.0 mLmL of water. (Assume a...

A solution contains 10.10 gg of unknown compound dissolved in 50.0 mLmL of water. (Assume a density of 1.00 g/mLg/mL for water.) The freezing point of the solution is -6.05 ∘C∘C. The mass percent composition of the compound is 38.70% CC, 9.74% HH, and the rest is OO. Part A What is the molecular formula of the compound? Express your answer as a molecular formula. nothing

When 1.31 g of an unknown non-electrolyte is dissolved in 50.0 g of acetone, the freezing...

When 1.31 g of an unknown non-electrolyte is dissolved in 50.0 g of acetone, the freezing point decreased to -94.1 degrees C from -93.4 degrees C. If the Kfp of the solvent is 2.4 K/m, calculate the molar mass of the unknown solute

An aqueous solution containing 34.1 g of an unknown molecular (non-electrolyte) compound in 159.9 g of...

An aqueous solution containing 34.1 g of an unknown molecular (non-electrolyte) compound in 159.9 g of water was found to have a freezing point of -1.5 ∘ C Calculate the molar mass of the unknown compound.

An unknown compound (152 mg) was dissolved in water to make 75.0 mL of solution. The...

An unknown compound (152 mg) was dissolved in water to make 75.0 mL of solution. The solution did not conduct electricity and had an osmotic pressure of 0.328 atm at 27°C. Elemental analysis revealed the substance to be 78.90% C, 10.59% H, and 10.51% O. Determine the molecular formula of this compound by adding the missing subscripts.

a solution is made by mixing 1.08 g of an unknown non-volatile non-electrolyte with 10.0 g...

a solution is made by mixing 1.08 g of an unknown non-volatile non-electrolyte with 10.0 g of benzene. The freezing point of pure benzene is is 5.5 degrees Celsius. The molal freezing point depression constant, Kf, for benzene is 5.12 degrees Celsius per molale. What is the value of the freezing point depression? What is the molality of the solution? What is the molar mass of the unknown?

A 1.31 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water...

A 1.31 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water and titrated with a a 0.496 M aqueous sodium hydroxide solution. It is observed that after 9.16 milliliters of sodium hydroxide have been added, the pH is 7.227 and that an additional 14.6 mL of the sodium hydroxide solution is required to reach the equivalence point. (1) What is the molecular weight of the acid? ____ g/mol (2) What is the value of Ka...

A solution containing 0.20 g of an unknown non-electrolyte in 2.50 g of cyclohexane was found...

A solution containing 0.20 g of an unknown non-electrolyte in 2.50 g of cyclohexane was found to freeze at 4.5 oC . What is the molar mass of the unknown substance?

A sample of sodium carbonate unknown is dissolved in 50.0 mL of water. After performing experiment...

A sample of sodium carbonate unknown is dissolved in 50.0 mL of water. After performing experiment 3, the student determines that the second equivalence point occurs at 32.86 mL of 0.1115 M HCl. What is the theoretical pH of the solution after 16.43 mL of HCl has been added. Report your answer to two significant figures. (HINT - refer to the lab procedure and posted notes for the correct way to approach this problem)

The osmotic pressure of a solution containing 1.44 g of an unknown compound dissolved in 175.0...

The osmotic pressure of a solution containing 1.44 g of an unknown compound dissolved in 175.0 mL of solution at 25 ∘C is 1.93 atm. The combustion of 32.88 g of the unknown compound produced 55.60 gCO2 and 22.76 gH2O. What is the molecular formula of the compound (which contains only carbon, hydrogen, and oxygen)? Express your answer as a chemical formula.

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