Calculate the molarity of a 14.17 mL solution which contains 453.76 mg of sodium sulfate (used in dyeing and printing textiles,

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

c. 75.7

i think that is the right

it takes on an electronical energy