problems: Given the following constants: lsig/100mL MW s grans rams of solute ar
ID: 191761 • Letter: P
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problems: Given the following constants: lsig/100mL MW s grans rams of solute are there in 300 mL of a 1,5% (wh) Naci solution! Show your work. How many moles of sucrose are there in 250 ml of a 1.5 M solution? Show your work How would you prepare 600mL of a 9% solution of NaCI Show your work. How many grams of solute are required to prepare 850 mL of a 0.2 M solution of gucose (w your work. gducose (Mw 180) sokution! Show How much solute 2) is required to prepare 150 ul of a 0.5 M solution of CuSO. (HW 5961) solution? Show your work The nutritional label on a bottle of soy sauce indicates that one serving (15 mL) of soy sauce contains 920 mg of sodium. What is the molarity of Na in this of this 15 ml serving (MW Na 230) A 100mL stock solution of sodium pyruvate was created at a concentration of I.75M. What is the percent concentration of this 1.75M sodium pyruvate solution (MW 110 g/mole)? Show your work. A solution has 0.5 mg/mL of the enzyme lysozyme. How much lysozyme is present in a. 5 mL of solution b. 100 of solution solution concentration (x-axis). A column graph with SEM bars is appropriate in this case. Please be sure that your figure description includes the number of measurements (N) and recognizes the SEM bars in the graph. Students: please turn in I graph next week with this worksheet Please graph average absorbance at 710 nm (-axis)Explanation / Answer
Ans. #1. Grams of solute = [NaCl] x Volume of solution
= (1.5 g/ 100 mL) x 300 mL
= 4.5 g
#2. Moles of sucrose = Molarity x Volume of solution in liters
= 1.5 M x 0.250 L ; [1 L = 1000 mL]
= (1.5 mol/ L) x 0.250 L ; [1 M = 1 mol/ L]
= 0.375 mol
#3. Required mass of NaCl = (9 g/ 100 mL) x 600 mL = 54.0 g
Dissolve 54.0 g NaCl in sufficient water to make the final volume to 600.0 mL. It is the 600 mL of 9% NaCl.
#4. Require moles of sucrose = 0.2 M x 0.850 L = 0.170 mol
Now,
Required mass of glucose = 0.170 mol x (180 g/ mol) = 30.6 g
#5. Moles of Na = Mass in gram / molar mass = 0.920 g / (23 g/mol) = 0.04 mol
Now,
Molarity of solution = moles of solute / Volume of solution in liters
= 0.04 mol / 0.015 L
= 2.67 M
#6. Moles of pyruvate = 1.75 M x 0.100 L = 0.175 mol
Mass of pyruvate = 0.175 mol x (110 g/ mol) = 19.25 g
Now,
% Pyruvate = (Mass of pyruvate / Volume of solution in mL) x 100
= (19.25 g /100 mL) x 100
= 19.25 % (w/v)
#7. #a. Amount of lysozyme = concertation x Volume of solution
= (0.5 mg/ mL) x 5 mL
= 2.5 mg
#b. Amount of lysozyme = (0.5 mg/ mL) x 0.100 mL
= 0.05 mg
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