When 0.60 grams of a non-electrolyte unknown compound is dissolved in 10 grams of water, the freezing point of the solution is -1.86o C. Given that the freezing point of pure water is 0o C and that the freezing-point depression constant for water is 1.86 o C/mol/kg, calculate the molecular mass (g/mol) of the unknown compound.
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When 0.60 grams of a non-electrolyte unknown compound is dissolved in 10 grams of water, the...
A solution contains 10.35 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 -3.26 ∘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?
A researcher dissolved 1.50 grams of an unknown compound in 75.0 grams of pure cyclohexane. The freezing point of the solution was measured to be 2.70 °C. Knowing the freezing point of cyclohexane is 6.50 °C and the freezing point constant is -20.2 °C kg/mol, calculate the molar mass of the unknown compound.
A solution contains 11.70 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 -7.01 ∘C. The mass percent composition of the compound is 38.70% C, 9.74% H, and the rest is O. What is the molecular formula of the compound? Express your answer as a molecular formula So far I got CH3O as my empirical, but I can't seem to figure out how...
A solution containing 1.00 g of an unknown non-electrolyte liquid and 9.00 g water has a freezing point of -3.33 oC. The Kf = 1.86 oC/m for water. Calculate the molar mass of the unknown liquid, in g/mol.
1a: 19.930 g of a non-volatile solute is dissolved in 395.0 g of water. The solute does not react with water nor dissociate in solution. Assume that the resulting solution displays ideal Raoult's law behaviour. At 90°C the vapour pressure of the solution is 521.11 torr. The vapour pressure of pure water at 90°C is 525.80 torr. Calculate the molar mass of the solute (g/mol). 1b: Now suppose, instead, that 19.930 g of a volatile solute is dissolved in 395.0...
A student is determining the molar mass of an unknown compound based on the following data: mass compound: 4.80 g mass solvent (water): 0.022 kg new freezing point: -2.50 oC freezing point depression constant (Kf): 1.86 oC/molal (Freezing point of pure water = 0 oC) 0.0786 g/mol 162.33 g/mol 0.00616 g/mol 293.26 g/mol
0.2650 g of a compound of unknown molecular mass were dissolved in 18.00 mL of a non-ionizing solvent with specific gravity of 0.7480. The pure solvent was determined to have a freezing point of 6.80°C. The freezing point of the solution was determined graphically. Trial one yielded a freezing point of 5.31°C for the solution. Trial two indicated the freezing point to be 5.23°C. Kf (solvent) = 12.8 C°/m a. Calculate the molecular mass of the unknown. b. If 5.50...
Determine the molar mass of an unknown non-electrolyte compound given the following data: 1.453 grams of the unknown compound is dissolved in 10.08 grams of water (Kf = 1.853 0C/m). The solution freezes at -1.47 0C. What is the molar mass of the unknown compound?
1h. A certain pure solvent freezes at 39.8°C and has a freezing point depression constant Kf = 0.777°C/m. What is the predicted freezing point (in °C) of a solution made from this solvent that is (1.90x10^0) m in a non-electrolyte solute? 1i. When (8.23x10^1) g of a non-electrolyte is dissolved in (5.2600x10^2) g of a solvent (with Kb = 0.416°C/m) the boiling point of the solution is 1.50°C higher than the boiling point of the pure solvent. What is the...
The freezing point of a solution that contains 1.00 g of an unknown compound, (A), dissolved in 10.0 g of benzene is found to be 2.17 oC. The freezing point of pure benzene is 5.48 oC. The molal freezing point depression constant of benzene is 5.12 oC/molal. What is the molecular weight of the unknown compound?