Calculate the molar mass (in g/mol) of an unknown 1:1 electrolyte if 0.45 g dissolved in 199.2 mL of water at 49.72 °C has an osmotic pressure of 34.4 mmHg.
R = 0.082058 L⋅atm⋅mol−1⋅K−1. 1.00 atm = 760 mmHg. Report your answer to THREE significant figures.
Calculate the molar mass (in g/mol) of an unknown 1:1 electrolyte if 0.45 g dissolved in...
3a. Calculate the molar mass (in g/mol) of an unknown 1:1 electrolyte if 0.482 g dissolved in 223.1 mL of water at 74.75 °C has an osmotic pressure of 54.4 mmHg. R = 0.082058 L⋅atm⋅mol−1⋅K−1. 1.00 atm = 760 mmHg. Report your answer to THREE significant figures. 3b. Calculate the required mass of an unknown nonelectrolyte (ℳ = 131.5599 g/mol) dissolved in 140.1 g of solvent that gives a solution that boils at 36.04 °C. The boiling point of the...
3a. Calculate the required mass of an unknown nonelectrolyte (ℳ = 125.7998 g/mol) dissolved in 129.7 g of solvent that gives a solution that freezes at 1.85 °C. The freezing point of the pure solvent is 4.44 °C and its Kf value is 1.87 °C/m. Report your answer to TWO places past the decimal. 3b. Calculate the molar mass (in g/mol) of an unknown nonelectrolyte if 0.613 g dissolved in 250.3 mL of water at 19.47 °C has an osmotic...
Colligative Properties 1a When a solution is made from 25.9 g of an unknown nonelectrolyte dissolved in 167.6 g of solvent, the solution boils at 63.24 °C. The boiling point of the pure solvent and its Kb are 59.02 °C and 2.09 °C/m, respectively. Calculate the molar mass of the unknown electrolyte in g/mol. Report your answer to TWO places past the decimal 1b When a solution is made from 39.5 g of an unknown nonelectrolyte dissolved in 109.1-2.47 °C....
2a Calculate the freezing point of a solution that is made from 39.9 g of a nonelectrolyte (ℳ = 142.4701 g/mol) dissolved in 186.3 g of solvent. The solvent freezes at 1.24 °C and its Kf value is 2.09 °C/m. Report your answer to TWO places past the decimal. 2b The solubility of carbon dioxide gas at 37.3 °C and a carbon dioxide pressure of 554 mmHg is 4.04 × 10−3 g/L. What is the Henry's Law constant in mol⋅L−1⋅atm−1?...
1.When a solution is made from 32.2 g of an unknown nonelectrolyte dissolved in 151 g of solvent, the solution boils at 83.44 °C. The boiling point of the pure solvent and its Kb are 79.31 °C and 4.47 °C/m, respectively. Calculate the molar mass of the unknown electrolyte in g/mol. 2. Calculate the molar mass (in g/mol) of an unknown nonelectrolyte if 0.898 g dissolved in 268.7 mL of water at 30.13 °C has an osmotic pressure of 68.1...
An unknown strong electrolyte in the form X2Y3 was found to have a molar mass of 262.85g/mol. If 3.58g of the compound is dissolved in enough water to make 175mL of solution at 298K, what is the osmotic pressure of the resulting solution (in atm)?
100. mL of solution and is found to have an osmotic pressure of 28.1 mmHg at 20°C. Calculate the molar mass of the compound. (1 atm 760 mmHg) 1 0.102 g of an unknown non electrolyte solute is dissolved in enough water to make A. 663 g/mol B. 0.872 g/mol C. 1.15 g/mol D. 727 g/mol E. 1.10 x 102 g/mol
A solution of an unknown nonelectrolyte is formed by adding 1.13 g in 250 mL of water. This generates an osmotic pressure of 0.46 atm at 25 °C. Given R = 0.0821 L-atm/K-mol, calculate the molar mass of the nonelectrolyte (to two significant figures). Please explain step-by-step.
A 2.337 g sample of magnesium carbonate (MgCO3) was dissolved in hydrochloric acid. The following reaction occurred at 27.1oC and a pressure of 784 mmHg: MgCO3(s) + 2HCl(aq) --> CO2(g) + H2O(l) + MgCl2(aq) Calculate the volume (in mL) of CO2 produced. Give your answer to the correct number of significant figures and in units of mL. Data: 1 atm = 760 mmHg, 0oC = 273 K, PV = nRT, R = 0.0821 L.atm.mol-1.K-1, Molar mass (g/mol) of Mg =...
A flask with a volume of 1.02 L contains 14.58 g of an unknown gas at 35.1 °C and 846 torr. What is the molar mass of the unknown in g/mol? Report your answer to TWO places past the decimal. Do not include units. The gas constant R = 0.082058 L⋅atm/mol⋅K and 1 atm = 760 torr.