Prelab 1 -BIO3126

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University of Ottawa *

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BIO3126

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Chemistry

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Jan 9, 2024

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Prelab 1-0 Exercise 1.1 1. What is the molarity of the methylene blue stock solution? (1 point) 0.26% m v of methylene ¿ = x gsolute 100 mL solvent x = 0.26 gof solute 0.26 gof solute × 1 mole 320 g = 8.215 × 10 − 4 mol 8.215 × 10 − 4 mol÷ 0.100 L = 8.215 × 10 − 3 M of methylene ¿ stock Therefore, molarity of methylene blue stock solution is 8.215 × 10 − 3 M ~ 8.215 mM 2. What dilution is required to prepare 5 mL of a 0.4mM starting solution? (1 point) C 1 V 1 = C 2 V 2 8.215 mM ×V 1 = 0.4 mM × 5.0 m L V 1 = 0.2462 mL≈ 0.25 mL 5 mL solution = 0.2462 mL + water water = 5 mL − 0.2462 mL = 4.7538 mL≈ 4.75 mL Therefore the dilution is 0.25:5 which means 1/20 dilution 3. What volumes of the methylene blue stock and water are required to prepare the above- described solution? (1 point) 0.2462 mL ≈ 0.25 mL for stock solution of methylene blue and 4.75 mL of water. 4. Complete the following table: (2 points) Tube Number Source tube number Vol. of water (mL) Vol. from source tube (mL) Dilution factor (X) Serial dilution factor (X) 2 1 1.2 4.8 1.25 1.25 3 2 3.0 2.5 2.2 2.75 4 3 1.5 2.0 1.75 4.81 5 4 1.0 1.6 1.63 7.82 6 5 1.0 1.5 1.67 13.06

Note: Dilutions must be expressed as a fraction where the numerator is 1. Ex. 1/5. Dilution factors must be expressed as a number followed by the symbol “X”. Ex. 5X. Exercise 1.2 5. What weight of sugar and salt are required to prepare 30 mL 30% solutions of each solute? (1 point) 30% = x g 30 mL × 100 x = 9.0 g 9.0 g of sugar and 9.0 g of salt are required to make a 30% (m/v) solution 6. What are the percent concentrations of glucose and sodium chloride in TSB? (1 point) Glucose concentration: 0.25% = 2.5 g 1000 mL × 100 Salt concentration: 0.5% = 5.0 g 1000 mL × 100 Total salt and glucose concentration = 0.25% + 0.5% = 0.75% (m/v) 7. How many grams of glucose and sodium chloride are present in 5 mL of 2X TSB? (1 point) There are 2.5 g of glucose and 5.0 g of sodium chloride in 1000 mL TSB, so in 1000 mL of 2X TSB, there will be 5 g of glucose and 10 g of sodium chloride. Grams of glucose: 5 g 1000 mL = x g 5.0 mL 25 g mL = 1000 x g mL x = 0.025 g Grams of sodium chloride: 10 g 1000 mL = x g 5.0 mL 50 g mL = 1000 x g mL x = 0.05 g

Therefore, in 5 mL of 2X TSB, there is 0.05 g of NaCl and 0.025 g of glucose. 8. Complete the following table: (1 point) Sugar conc. (%) Vol. stock soln. (mL) Vol. of water (mL) Vol. 2X TSB (mL) Dilution factor (X) 0.5 0.083 4.917 5.0 60 5 1.583 3.417 5.0 3.16 10 3.250 1.750 5.0 1.54 15 4.917 0.083 5.0 1 Salt conc. (%) 0.5 0.000 5.000 5.0 0 5 1.500 3.500 5.0 3.33 10 3.167 1.833 5.0 1.58 15 4.833 0.167 5.0 1 Sample calculation for sugar conc. (%) at 0.5% 0.05% = x 10 mL × 100 x = 0.05 g 0.05 g of sugar in the final solution. Since there is 0.025 g of glucose in the 5 mL of 2X TSB, then, 0.05 g − 0.025 g = 0.025 g There should be 0.025 g of sugar is needed from the 30 mL stock solution Since we have 9.0 g of sugar needed to make the stock solution (30% m/v) of 30 mL, 9 g 30 mL = 0.025 g x x = 0.083 mL Vol. of water (mL): 10 mL – 0.05 mL – 0.083 mL = 4.917 mL Therefore, the vol. of solution is 0.083 mL and the volume of water is 4.917 mL.

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dilution factor = 5 mL 0.083 mL = 60 Exercise 1.3 9. Use a diagram to illustrate how 10 -4 , 10 -6 and 10 -8 dilutions in water of the soil suspension in a final volume of 10 mL will be prepared. Note you only have three tubes available. (1 point) For 10 -4 dilution: 1 10000 = x mL 10 mL 10 mL 10000 = x 1 × 10 − 3 mLof stock = x 10.0 mL - 0.001 mL = 9.99 mL of water 0.001 mL or 1  L of stock solution and 9.99 mL of water is needed to make a dilution of 10 -4 in Tube 1. For 10 -6 dilution: 1 100 = x mL 10 mL 10 mL 100 = x 0.1 mLof Tube 1 = x 10.0 mL – 0.1 mL = 9.90 mL of water 0.1 mL of Tube 1 and 9.90 mL of water to is needed to make a dilution of 10 -6 dilution in Tube 2. For 10 -8 dilution: 1 100 = x mL 10 mL 10 mL 100 = x 0.1 mLof Tube 2 = x 10.0 mL – 0.1 mL = 9.90 mL of water 0.1 mL of Tube 2 and 9.90 mL of water is needed to make a dilution of 10 -8 dilution in Tube 3.