Calculate the mass of KCl required to prepare 250. mL of 0.250 M solution. 3. Calculate the volume of 0.30 M KCl solution that contains 6.00 g of KCl. 6.00 g x 1 mole x 1 L = 0.27 L 74.6 g 0.30 mol 4. Calculate the volume of 0.250 M H2SO4 that contains 0.250 g H2SO4.
of mixture Metal C 25.605g 24.6mL 25.2°C 100.5°C 28.7°C Calculations: Show your work and write a short explanation with each calculation. Part I: 1. Calculate the energy change (q) of the surroundings (water). We can assume that the specific heat capacity of water is 4.18 J/ (g · °C), and the density of water is 1.00 g/mL. (4 points) q = m × c × Δt Given: q=?
CHEM 1412 SAMPLE FINAL EXAM PART I - Multiple Choice (2 points each) _____ 1. In which colligative property(ies) does the value decrease as more solute is added? A. boiling point B. freezing point and osmotic pressure C. vapor pressure D. freezing point and vapor pressure _____ 2. What is the molarity of a solution prepared by dissolving 25.2 g of CaCO3 in 600 mL of solution? A.
Mixed Solutions: If 1 mmole of glucose (180mg=1mOsm) and 1 mmole of NaCl (58mg=2mmOsm) are put into a beaker and distilled water added to make 1 liter, the osmolarity is 3 mOSm/L. OSMOSIS CALCULATIONS: 1. Calculate the number of grams of NaCl needed to prepare 100 mL of a 280mOsm.L Solution. (280mosm/l) *(1L/1000mL)*(100mm/1)*(29mg/1mosm)*(1g/1000mg)=0.812g 2. Calculate the number of grams of glucose needed to prepare 100mL of a 280 msm/L glucose solution.
There will have some error. 2) A volatile liquid was allowed to evaporate in a 43.298 g flask that has a total volume of 252 ml. the temperature of the water bath was 100˚C at the atmospheric pressure of 776 torr. The mass of the flask and condensed vapor was 44.173 g. calculate the molar mass of the liquid. T = 273 + 100 = 373 V = 252 mL = 1 L / 1000 mL = 0.252 L P = 776 Torr R= 0.0821 mass of 44.173 - 43.298 g = 0.875g moles of gas = PV / RT = 776 x .252 / 62.363 x (273+100) =0.00841 moles molar mass = 0.875g / 0.00841 moles = 104.1 g/
Eighth, and last, I found molecular weight of sugar and diluted it. Data Tables and Observations: Data Table 1: Length measurements. | Object | Length (cm) | Length (mm) | Length (m) | CD or DVD | 5.2 | 52.0 | .052 | Key | 12.0 | 120.0 | .120 | Spoon | 18.6 | 186.0 | .186 | Fork | 21.0 | 210.0 | .210 | Data Table 2: Temperature measurements. | Water | Temperature (°C) | Temperature (°F) | Temperature (K) | Hot from tap | 46 | 114.8 | 319.15 | Boiling | 100 | 212 | 373.15 | Boiling for 5 minutes | 100 | 212 | 373.15 | Cold from tap | 18 | 64.4 | 291.15 | Ice water – 1 minute | 0 | 32 | 273.15 | Ice water – 5 minutes | 0 | 32 | 273.15 | Data Table 3: Mass measurements. | Object | Estimated Mass (g) | Actual Mass (g) | Actual Mass (kg) | Pen or pencil | 5 | 5.4 | .0054 | 3 Pennies | 6 | 7.6 | .0076 | 1 Quarter | 5 | 5.7 | .0057 | 2 Quarters, 3 Dimes | 15 | 18.2 | .0182 | 4 Dimes, 5 Pennies | 15 | 21.6 | .0216 | 3 Quarters, 1 Dime, 5 Pennies | 25 | 31.9 | .0319 | Key | 10 | 9.2 | .0092 | Key, 1 Quarter, 4 Pennies | 20 | 24.9 | .0249 | Data Table 4: Liquid measurements.
Pre-Lab Questions 1. A determination of the molar mass of methyl alcohol (CH3OH) yielded the following data. Temperature of boiling water bath 99.5o C Barometric pressure 738 mm Hg Temperature of room temperature water 24.0o C Density of room temperature water 0.9973 g/mL Trial 1 Mass of empty pipette 1.557g Mass of piet and condensed methyl alcohol 1.571g Mass of pipet and water 16.001g Mass of condensed methyl alcohol 0.014g Mass of water in filled pipet 14.444g Volume of pipet 14.483mL Molar mass of methyl alcohol (experimental) 28.69 g/mol Molar mass of methyl alcohol (theoretical) 32.05 g/mol Using the data, fill in the rest of the table. The volume of the pipet is equal to the volume of water inside the pipet. Use the relationship of mass and density to determine this volume.
Reactions Lab David Vaghari INSTRUCTOR: Dr. Chernovitz Monday, July 23, 2012 Oxygen Production Introduction In this lab, potassium chlorate will be decomposed producing oxygen gas and potassium chloride. The hypothesis is that the reaction will yield 3.916 grams of oxygen gas. Materials Test tube 10 grams potassium chlorate Bunsen burner Procedure Step 1. Obtain a test tube, place a 10 gm of potassium chlorate. Step 2.
Preparation 1. Sodium bicarbonate solution: Mix 1 g of sodium bicarbonate in 100 mL of distilled or deionized water. 2. Neutral red solution: Mix 0.2 g of neutral red in 1 L of distilled water. 3.
Used |B.P/M.P |Density |Structure | |Cyclohexane |84.16amu |“25ml” |80.74° C |0.779g/mL (l) |[pic]2 | |Benzoic Acid |122.12amu |1.00g |250°C |1.27g/ml |[pic]3 | Procedure/Observations 1.) Filtration/Recrystallization: Benzoic acid (1.00g) was weighed in a beaker (250ml). After adding distilled water (100ml) the beaker was heated with a Bunsen burner until the solid dissolved. After the solution sat to cool some solid recrystallized. The liquid was cooled in