CHEM 4025 Study Guide - Final Guide: Central Board Of Secondary Education, Molality, Molar Mass

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17 Aug 2021

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Cameron University

Department

Chemistry

Course

CHEM 4025

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Solutions are the homogeneous mixtures of two or more than two components: binary solution: A solution having two components is called a binary solution: components of a binary solution. It is the amount of solute in given amount of solution: mass by volume percentage (w/v): Mass of the solute dissolved in 100 ml of solution: molality (m) is the number of moles of solute present in 1kg of solvent. Mass of solvent in kilograms: molarity (m) is the number of moles of solute present in 1l of solution. Volume of solution in litres: normality is the number of gram equivalent of solute dissolved per litre of solution. It is the maximum amount that can be dissolved in a specified amount of solvent at a specified temperature. It is a solution in which no more solute can be dissolved at the same temperature and pressure.

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Questions: THANK YOU IN ADVANCE

1. WHAT IS THE TERM FOR THE CONCENTRATION EXPRESSION THAT RELATES THE MOLES OF SOLUTE DISSOLVED IN EACH LITER OF SOLUTION?

A. MOLARITY

B. MASS/MASS PERCENT

C. MOLALITY

D. PARTS PER MILLION

E. NONE OF THESE

2. WHAT IS THE TERM FOR A SOLUTION CONTAINING THE MAXIMUM SOLUTE THAT CAN DISSOLVE AT A GIVEN TEMPERATURE?

A. SATURATED

B. SUPERSATURATED

C. UNSATURATED

D. CONCENTRATED

E. NONE OF ABOVE

3. WHICH OF THE FOLLOWING ILLUSTRATES THE LIKE DISSOLVES LIKE RULE FOR TWO LIQUIDS?

A. a POLAR SOLUTE IS MISCIBLE W A NONPOLAR SOLVENT

B. A NONPOLAR SOLVENT IS MISCIBLE WITH A POLAR SOLVENT

C. A POLAR SOLUTE IS IMMISCIBLE WITH A POLAR SOLVENT

D. A NONPOLAR SOLUTE IS MISCIBLE WITH A NONPOLAR SOLVENT

E. NONE OF ABOVE

4. APPLY THE LIKE DISSOLVES LIKE RULE TO PREDICT WHICH OF THE FOLLOWING LIQUIDS IS MISCIBLE WITH WATER

A. ETHANOL, C2H5OH

B. TOLUENE, C7H8

C. CARBON TETRACHLORIDE, CCL4

D. ALL OF THESE

E. NONE OF THESE

5. WHICH OF THE FOLLOWING INCREASES THE RATE OF DISSOLVING FOR A SOLID SOLUTE IN A SOLVENT?

A. HEATING THE SOLUTION

B. GRINDING THE SOLUTE

C. STIRRING THE SOLUTION

D. ALL OF THE ABOVE

E. NONE OF THE ABOVE

6. WHAT IS THE MASS OF SOLUTE DISSOLVED IN 10.0 G OF A 5.00% SUGAR SOLUTION?

A. 0.900g

B 9.50 g

C 0.500g

D 0.180 g

E 10.0 g

THANK YOU!!!!!!!!!!!!!!!!!!!!!!!!

Please help me solve for the molar mass of sample 1 and 2 and the molecular formula

Experiment 1084-03: Determination of M.M. by Freezing Point Depres sion Measure the freezing-point depressions solute whose empirical formula is known Pool the class hat occur with solutions containing mass of the solute and its Background Information However, as soon as the solution begins to freeze, Adding a solute to a pure, non-volatile lquid ature will remain constant untl all the (solvent) lowers the vapor pressure of the solvent raises its boiling point, and lowers (depresses) its freezing point. The extent to which these are aftected depends on the relative number of solute particlos in the solution. Properties of solutions that depend only on the number of solute molecules in a given quantity of solvent are called colligative properties. Such cyclohexane turns into solid. Once the substance again. The freezing point is the temperature where it stays constant, Le. the liquld phase and the sold phase of the cyclohexane coexists (Figure 3-1) boiling point elevation and osmotic pressure. They do NOT dopend on the nature of the solute, but DO depend on the nature of the solvent. As a consequence, colligative properties can be used to determine an unknown molar mass for a solute. In this experiment, we will be using cyclohexane (d " 0.779 gmL, K, . 20.5 oC/m) as the solvent to determine the molar mass of a white solid with an empirical formula of C,HC. The freezing point depression, Î7, is the difference between the freezing point of the pure solvent, TP, and the freezing point of the solution (solvent and solute). T As mentioned above, thits quantity is proportional only to the molal concentration of the solute, m Figure 3-1. Cooling Curves The proportionality constant, K, is called the freezing-point-depression constant. The K value of 20.5 C/m for Cyclohexane is one of the largest values one can find. Molal concentration is defined as the number of moles of substance in 1 kg of solvent. Now we can calculate the molar mass of a solute from ATã K,, the mass of the The freezing point of a solution in cyclohexane can be obtained in a similar way. Cooling results in an initial rapid decrease in the temperature until freezing begins as seen in the cooling curve of the pure solvent. However, the temperature of the solution does not remain constant while being frozen. As the cyclohexane in the solution treezes, cyclohexane is taken away from the solution and the concentration of the solute becomes higher In the liquid portion. Thus the molal concentration of the solute increases and the freezing point of the solute and the mass of the solvent. remaining liquid becomes lower. The result is a steadily decreasing freezing temperature until the solution is completely frozen. When all the solution In order to determine ATP, we will measure the freezing point of pure cyclohexane and the freezing point of a solution of the solute in is frozen, the temperature decreases more rapidly To evaluate the freezing point of pure cyclohexame, you will cool a sample of this substance, and measure the temperature as a The freezing point of the solution is the temperature at which the solution starts to freeze. portion represents the concentration of the starting function of time. As cooling begins, theOnly at this point, the concentration of the liquid temperature of the cyclohexane decreases rapidly.

7. Complete the following table:

Element Moles Mass Num

7. Complete the following table:

Element Moles Mass Number of molecules

13. It is the component that is in a solution, in smaller quantity.

a. Solution

b. Solvent

c. Solvent

d. Soluto

14. It is the component that is in a solution in greater quantity.

a. Solution

b. Solvent

c. Solvent

d. The answers b and c are correct.

15. The quantitative expression of the dissolved amount of solute in a quantity determined solvent is known as:

a. Solution

b. Volume

c. Concentration

d. Stoichiometry

16.Express the moles of solute among the liters of solution

a. Molality

b. Molarity

c. % weight by volume

d. % weight by weight

17.Express the moles of solute between kg of solvent

a. Molarity

b. ppm

c. ppt

d. Molality

18. What is the molarity of a solution prepared by dissolving 7.50 g of nitrate of magnesium, Mg (NO3) 2 in enough water to prepare 25.0 mL of solution?

a. 10.3 M

b. 5.6 M

c. 6.5 M

d. 2.03 M

19. How many grams of sodium chloride, NaCl are needed to prepare 125 mL of 0.037 M NaCl solution

a. 1.021 M

b. 0.268 M

c. 0.201 M

d. 0.500 M

20. Calculate the molality of a solution containing 12.5 g of ethylene glycol (C2H6O2) dissolved in 135 g of water.

21. A solution of ethylene glycol contains 21.2 g of ethylene glycol (C2H6O2) in 85.4 mL of water. Calculate the freezing point and the boiling point of the solution. (Assume the water density is 1.00 g / mL)

22. At what volume should you dilute 75 mL of a 10.0 M solution of H2SO4 to obtain a 1.75 M solution of H2SO4.

CHEM 4025 Study Guide - Final Guide: Central Board Of Secondary Education, Molality, Molar Mass

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