38-39: consider the two standard curve graphs (below) generated from the Determi

Question

38-39: consider the two standard curve graphs (below) generated from the Determining Salt Concentrations experiment: Density vs NaCl concentration 1.020 1.015 010 2.8 29 .0 31 32 33 35 % Nad Concentration Conductivity vs NaCI concentration 2.20 2.15 2.10 . 2.8 2.9 30 31 32 33 34 3.5 34 % NaCl concentration 38. An unknown concentration of sodium chloride solution (represented by the point on the graphs) was determined to have a density of 1.016 g'ml. Referring to the graphs above, what is the numerical conductivity value of the unknown solution? A. 1.013 B. 3.15 C. 2.16 E. 2.22 39. Given the two graphs above, which of the following statements are true? (Mark all that apply) A. As percent NaCI concentration increases, conductivity increases. B. As percent NaCI concentration decreases, density increases C. As percent NaCI concentration increases, conductivity decreases. D. As percent NaCl concentration increases, density increases. E. As percent NaCI concentration decreases, conductivity increases. 40, The nutrition label on the back of a can of soup indicates there are 611 mg of sodium per ½ cup. What is the salt concentration of the soup in % (gimL)? There are 8 ounces in 1 cup. There are 2957 mL in 1 ounce. A. 1.33% B. 6.61% C. 15.1% D. 2.67% E. 0.517% END OF EXAM You must turn in ALL testing materials to receive a gra

Explanation / Answer

38.From first graph we nitice that the coumpound with density 1.016 g/cm^3 has % NaCl of 3.15 and for % NaCl of 3.15 in graph 2 we have conductivity of 2.16. So option C is correct.

39. From both the grapf we see that both density of solution and conductivity varies directly with % NaCl concentration.So Option A and D both are true.

40.Mass of sodium in 1/2 cup is 611 mg=0.611 grams

There are 8 ounces in 1 cup and as the cup is half filled so there are 4 ounces and each ounce volume is equal to 29.57 mL, sofor 1/2 cup the volume is 4x29.57 so total volume is 118.28 mL.

So concentration is given by: mass/volume =0.611/118.28=0.00517 g/mL

so % conc. is given by 0.00517 x 100= 0.517 % NaCl concentration. so option D is correct answer.

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