Question

commercial bleaching solution contains 5.25% by mass of NaClO in water.It has density of 1.08 g/mL. Calculate the molarity of the solution. (HINT:molar mass of NaClO 74.4 g/mol.)

Answer 1

5.25% NaClO means 5.25 g NaClO in 100 g solution or
5.25 g NaClO/100 g.
mols NaClO = grams/molar mass = ? estimated 0.08
What is the volume of 100 g of solution? That's mass = volume x density and volume = mass/density = 100/1.08 = estimated 90 mL
Molarity = mols/L solution. Plug in mols and L solution and you have it. Remember those numbers I have are estimates.

Answer 2

can someone help with this


1. What is the voltage for: Cu2+ + Zn(s) → Cu(s) + Zn2+ at 298 K if [Cu2+] = 0.15 M and [Zn2+] = 4.0 M? E°cell = 1.20 V.



2. A 25.0 mL of aliquot of a well-shaken and filtered sample of river water is pipetted into an evaporating dish. The sample was heated to dryness. Determine the TDS content, express (a) in ppt and; (b) ppm using the

following data:

• Mass of evaporating dish = 24.44 g

• Mass of water sample + evaporating dish = 49.44g

• Mass of dried sample + evaporating dish = 25.37 g



3. A 25.0 mL of aliquot of a well-shaken of river water is pipetted into an evaporating dish. The sample was heated to dryness. Determine (a) TS and (b) TSS and express in terms of ppm using the following data:

• Mass of evaporating dish = 25.25 g

• Mass of dried sample + evaporating dish = 28.14 g



4. Solve the missing data:

Sample volume (mL) = 250

Buret Reading, initial (mL) = 5.57

Buret Reading, final (mL) = 25.25

Volume of Na2S2O3 dispensed (mL) = ________

Average molar concentration of Na2S2O3 (mol/L) = 0.025

Moles of Na2S2O3 dispensed (mol) = __________

Moles of I3- reduced by S2O32- (mol) = __________

Moles of O2 (mol) = __________

Mass of O2 (mg) = ___________

Dissolved oxygen, ppm O2 (mg/L) = ___________