A chemist is given a nonelectrolyte white powder for analysis. 38.7 g of this substance is dissolved in 100.0 g of water. It is determined that the resulting solution freezes at -4.0°C. Additionally, the elemental composition of the substance is identified and found to be 40.0% C, 6.7% H and 53.3 % O by mass. Find the molecular formula of this unknown white powder using these experimental evidences (K of water: 1.86°C/m, C: 12 g/mol, H: 1 g/mol, O: 16 g/mol).

A chemist is given a nonelectrolyte white powder for analysis. 38.7 g of this substance is dissolved in 100.0 g of water. It is determined that the resulting solution freezes at -4.0°C. Additionally, the elemental composition of the substance is identified and found to be 40.0% C, 6.7% H and 53.3 % O by mass. Find the molecular formula of this unknown white powder using these experimental evidences (K of water: 1.86°C/m, C: 12 g/mol, H: 1 g/mol, O: 16 g/mol).

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

You are a forensic chemist and vou are given a white powder to analyze. You take a 0.25- g sample of this powder and dissolve it in 4.0 g of benzene. You determine that the freezing point of this solution is 3.9°C. Can you conclude that the compound is cocaine (C17H21NO4)? What assumptions did you make in the analysis?
A certain substance X melts at a temperature of - 7.4 °C. But if a 350. g sample of X is prepared with 37.7 g of urea ((NH2),CO) dissolved in it, the sample is found to have a melting point of - 11.0 °C instead. Calculate the molal freezing point depression constant of X. Be sure your answer has the correct number of significant digits. I C mol" kg 1 ·kg
A 12.7 g substance was dissolved in 2.00 g of solvent. The molar mass of the unknown substance is 169.7 g/mol. Calculate the freezing point depression. Kfof the substance = 6.0 K kg/mol.
At a given temperature, you have a mixture of ethanol and acetone at 19.5 °C. The vapor pressure of pure ethanol at 19.5 °C is 42.28 mmHg and the vapor pressure of pure acetone at 19.5 °C is 180.5 mmHg. The mole fraction of ethanol in the solution is 0.819. Assuming ideal behavior, calculate the mole fraction of acetone in the vapor above the solution, in mmHg.
2.00 g of an unknown compound is mixed to form a homogeneous solution with 75.00 g of benzene. This reduces the freezing point from 5.53 to 4.90 °C. What is the molar mass of the compound? AT;= – Kf × m; K5= 5.12 for benzene
15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at – 0.34 °C. What is the molar mass of the substance? (Kf for water = 1.86 K kg mol-1)
please explain
sited [Review Topics] [References] Use the References to access important values if needed for this question. The boiling point of diethyl ether, CH3 CH2 OCH2 CH3, is 34.500 °C at 1 atmosphere. K, (diethyl ether) = 2.02 °C/m In a laboratory experiment, students synthesized a new compound and found that when 10.14 grams of the compound were dissolved in 270.0 grams of diethyl ether, the solution began to boil at 34.714 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound? g/mol Submit Answer ph_08-30-2....pdf Retry Entire Group a a 9 more group attempts remaining C Previous Next 07 Show all 11:14 PM 9/12/2022
1. A solution was obtained by dissolving 25 grams of nonelectrolyte in 1000 grams of water. The freezing temperature for this solution equals as to - 1.55 °C. Determine the molar mass of dissolved substance. Cryoscopic constant for water equals to 1.86 K·Kg/mol.
The vapor pressure of pure liquid CS2 is 0.3914 atm at20°C. When 40.0 g of rhombic sulfur is dissolved in 1.00kg of CS2, the vapor pressure of CS2 decreases to 0.3868atm. Determine the molecular formula for the sulfur molecules dissolved in CS2.
Students are investigating a solvent for use in freezing point depression experiments. They measured the freezing point of the solvent as −12.6 ⁰C. The solution they made by dissolving 0.484 g of ethanol (46.07 g/mol) in 8.254 g of the solvent has a freezing point of -18.3 ⁰C. Determine the freezing point depression constant, Kf, for the solvent.
At 25.0 °C, the vapor pressure of pure carbon tetrachloride (CCl4) is 143. torr and the vapor pressure of pure chloroform (CHCl3) is 199 torr. The molar mass of CHCl3 is 119.38 g/mol and the molar mass for CCl4 is 154.8 g/mol.a. Do you expect the vapor pressure of a pure solvent to differ from the vapor pressure of a solution of that solvent? Why? Use the Raoult’s Law to explain.b. Calculate the vapor pressure in torr of a solution prepared by mixing 25.00 g of CCl4 and 20.00 g of CHCl3. c. Suppose you have a solution which contains CCl4 and CHCl3. Calculate the mole fraction of CHCl3 (XCHCl3) in a solution whose vapor pressure is 185. torr. Please report XCHCl3 with 3 sig figs. Recall the at 25.0 °C, the vapor pressure of pure CCl4 is 143. torr and the vapor pressure of pure chloroform (CHCl3) is 199 torr.