Answers are confirmed correct 1. [Unit conversions] Compute the concentrations f

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Answers are confirmed correct

1. [Unit conversions] Compute the concentrations for the solutions given below in terms of percentage by weight (wt%), parts per million (ppm) and molarity (M). (a) 0.003 g of ethanol in 106 mg of water (b) 36 g of HCl in 64 cm3 of water (c) 34 g of ammonia in 2 kg of water Ans: 3x104 wt%; 3 ppm; 6.51x105 M] [Ans: 36 wt%; 360000 ppm; 10.7 M] Ans: 1.67 wt%; 16716 ppm; 0.977 M] Supplementary information: Substance Density (kg m3) Molar mass (g mot) 18.02 36.46 17.03 46.07 1000 1268 817 789 Percentage by weight (wt%)-[mass of solute (g) / mass of solution (g)] x 100 Parts per million (ppm or ppmm) -[mass of solute (g) /mass of solution (g)x 106 Molarity (M) number of moles of solute (mol) / volume of solution (L)

Explanation / Answer

a)

0.003 g ethanol in 10^6 mg of water

weight % = mass of solute / mass of solution ) x 100

               = (0.003 / 1000) x 100

weight % = 3 x 10^-4 %

ppm = 3 ppm

Molarity = moles / volume

moles of ethanol = 0.003 / 46.07 = 6.51 x 10^-5 mol

volume of water = 1 L

Molarity = 6.51 x 10^-5 M

b)

mass of HCl = 36 g

moles of HCl = 36 / 36.46 = 0.9874 mol

mass of water = 64 g

wt % = 36 / 64 + 36 ) x 100

wt % = 36 %

ppm = 360000

volume of solution = 0.064 + 0.02839 = 0.0924 L

Molarity = 0.9874 / 0.0924 = 10.7 M

c)

wt % = (34 / 2000 + 34 ) x 100

         = 1.67 %

ppm = 16716 ppm

moles of NH3 = 34 / 17 = 1.996

Molarity = 1.996 / 2 = 0.977

Molarity = 0.977 M

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