Question

A solution at 25°C has a concentration of the permanganate ion, MnO₄^- (aq) = 2.38 mg/L. The solution has a density = 1 kg/L.

What is the concentration of MnO₄^- (aq) in units of µg/L?

What is the concentration of MnO₄^- (aq) in units of g/L?

What is the concentration of MnO₄^- (aq) in units of moles/L?

What is the concentration of MnO₄^- (aq) in units of millimoles/L?

What is the concentration of MnO₄^- (aq) in units of micromoles/L?

What is the concentration of MnO₄^- (aq) in units of parts per million (ppm)?

What is the concentration of MnO₄^- (aq) in units of parts per billion (ppb)?

What is the concentration of MnO₄^- (aq) in units of moles/kg solution?

Answer 1

= 2.38X II ANSWER: given [Monocoq)) = 2:38mg/L. ů come of Miron (19.) = 2.38 mg / 1 m 2 = 100 syll] [mmor leg)) mg/l = 2:36 mg/l [mro; 109) ug|1= 2:38X10ugL w my IL to gli Imple = 103g11. [mmo , copyTagle = 2:30 mgle [mov a ces), I2 = 2:38%100g/. i mg/L moles / for this we have to calculate 1 mg IL g/L molus IL. cone of mnou (99) in g/L = 2.38 x 10-3 g/L. for moles / L = 238 X 10 -3 = 2.38X10-3 molar mass of mnou 118.936 = 0.2210-4 moles 12 liv for cone. of (mnou] in millimoles IL = 0.qx 104 moly IL = 0.2X10~Yx10.8 milli moles / (1 millimolella 10-3 molely

nc. ( for conc. of (mnou] in millimeler / 30:2X10-' milli moly.L= 0.02 milli moles L. for conc. of Im now] in micromoles / = 0.2x10-4 moles /L 0.2X10" x 100 micromeler / mole/1 = 106 mitromoles (2) = 20 micro moles IL (vi) we have the conc. of [mmoy - ] = 2.38 mg IL Img I L'=ippm cone of (mnou] in ppm = 2.38 ppm. - (vii) for ppb; ImgIL = loro ppb conc. of [mrow. I in ppb = 22:38x 15°ppbo. Vil for conc. of MnO4 (aq) in moles Kg solution to find; divide the moles Il by density; we get: [mmoylaq)] in mols )1 = 0.2x10 4 molesle from eart it) Im nou lag) in molus (kg = 0.2x10-"moles / [monos e) in z = 0.210-4 roling om 0.2 x 10 moles / kg