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marxusalexander6552
14.07.2020 •
Chemistry
Calculate the molarity of a 14.17 mL solution which contains 453.76 mg of sodium sulfate (used in dyeing and printing textiles,
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Ответ:
0.226 mol/L
Explanation:
The following data were obtained from the question:
Volume = 14.17 mL
Mass of sodium sulphate, Na2SO4 = 453.76 mg
Molarity =..?
Next, we shall determine the number of mole in 453.76 mg of Na2SO4. This is illustrated below:
Molar mass of Na2SO4 = (2x23) + 32 +(16x4) = 142g/mol
Mass of Na2SO4 = 453.76 mg = 453.76×10¯³ g
Mole of Na2SO4 =..?
Mole = mass /Molar mass
Mole of Na2SO4 = 453.76×10¯³ / 142
Mole of Na2SO4 = 3.20×10¯³ mole
Finally, we shall determine the molarity of Na2SO4 solution as follow:
Mole of Na2SO4 = 3.20×10¯³ mole
Volume = 14.17 mL = 14.17/1000 = 1.417×10¯² L
Molarity =?
Molarity = mole /Volume
Molarity = 3.20×10¯³ / 1.417×10¯²
Molarity = 0.226 mol/L
Therefore, the molarity of the sodium sulphate, Na2SO4 solution is 0.226 mol/L
Ответ:
Democritus!
Explanation:
Brainliest!? Pl