Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Which of the following are correct equivalentvalues ? For any solutions specifie

ID: 680133 • Letter: W

Question

Which of the following are correct equivalentvalues ? For any solutions specified, assume they havethe density of water.
a) 0.050 mg/L = 50 ppm b) 3.0 ppm = 3.0 g solute / 106 gsolution c) 25 g/L = a weight percent of 0.25 d) 25 ppm = 2.5 × 105 g solute/ 1 g solution e) 3.0 ppm = 3.0 mg solute / kg solution Which of the following are correct equivalentvalues ? For any solutions specified, assume they havethe density of water.
a) 0.050 mg/L = 50 ppm b) 3.0 ppm = 3.0 g solute / 106 gsolution c) 25 g/L = a weight percent of 0.25 d) 25 ppm = 2.5 × 105 g solute/ 1 g solution e) 3.0 ppm = 3.0 mg solute / kg solution

Explanation / Answer

(a)   0.050 mg/L = 50 ppm 1 ppm = 1 mg / L So this is false (b) 3.0 ppm = 3.0 g solute / 106 g solution ppm = (Mass of solute / mass of solution) *106 So this is true. (c)    25 g/L = a weight percent of 0.25 Since density = 1.00g/ mL 25 g/L = 25 g / kg            =25 per 1000 g Mass percent = ( 25 g/ 1000 g) * 100                      =2.5 % So this is false (d)    25 ppm = 2.5 × 105 g solute / 1 gsolution ppm = (Mass of solute / mass of solution) *106         = (2.5 ×105 g / 1 g) * 106         = 25 ppm so this is true. (e) 3.0 ppm = 3.0 mg solute / kg solution ppm ={ (3 * 10^ -3 g ) /  10^3 g} * 10^6         = 3.0 ppm so this is true. 25 g/L = 25 g / kg            =25 per 1000 g Mass percent = ( 25 g/ 1000 g) * 100                      =2.5 % So this is false (d)    25 ppm = 2.5 × 105 g solute / 1 gsolution ppm = (Mass of solute / mass of solution) *106         = (2.5 ×105 g / 1 g) * 106         = 25 ppm so this is true. (e) 3.0 ppm = 3.0 mg solute / kg solution ppm ={ (3 * 10^ -3 g ) /  10^3 g} * 10^6         = 3.0 ppm so this is true.           
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote