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.)
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.
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) = ___________
To calculate the molarity of the solution, we need to know both the amount of solute and the volume of the solution.
First, let's calculate the amount of NaClO in grams. Since the solution contains 5.25% of NaClO by mass, we can assume that 100 grams of the solution will contain 5.25 grams of NaClO.
Next, we need to find the volume of the solution in liters. We are given the density of the solution, which is 1.08 g/mL. Thus, the mass of 1 mL of the solution will be 1.08 grams.
To convert the mass of the solution to volume, we can use the following conversion:
1 g = 1 mL
Therefore, the volume of the solution in liters will be:
100 grams of solution * (1 mL / 1.08 grams) * (1 L / 1000 mL) = 0.9259 L
Now, let's calculate the molarity of the solution using the given molar mass of NaClO (74.4 g/mol).
Molarity (M) is defined as the number of moles of solute (NaClO) per liter of solution (L).
To find the number of moles of NaClO, we can use the following formula:
moles = mass (g) / molar mass (g/mol)
moles of NaClO = 5.25 g / 74.4 g/mol = 0.07065 mol
Finally, we can calculate the molarity using the formula:
Molarity (M) = moles / volume (L)
Molarity = 0.07065 mol / 0.9259 L = 0.0762 M
Therefore, the molarity of the solution is 0.0762 M.
To calculate the molarity of the solution, we need to determine the number of moles of NaClO present in the solution and then divide it by the volume of the solution in liters.
First, let's calculate the mass of NaClO in 1 mL of the solution. Given that the solution has a density of 1.08 g/mL, the mass of 1 mL of the solution will be:
Mass of solution (1 mL) = Density × Volume
Mass of solution (1 mL) = 1.08 g/mL × 1 mL
Mass of solution (1 mL) = 1.08 g
Since the solution contains 5.25% NaClO by mass, we can calculate the mass of NaClO in 1 mL of the solution:
Mass of NaClO (1 mL) = 5.25% of Mass of solution (1 mL)
Mass of NaClO (1 mL) = 5.25% of 1.08 g
Mass of NaClO (1 mL) = 0.0525 g
Next, we can convert the mass of NaClO in 1 mL to moles. We can use the molar mass of NaClO, which is given as 74.4 g/mol:
Number of moles of NaClO = Mass of NaClO / Molar mass
Number of moles of NaClO = 0.0525 g / 74.4 g/mol
Number of moles of NaClO = 0.000706 mol
Now, we need to convert the volume of the solution from milliliters to liters. Since there are 1000 mL in 1 L, the volume of the solution in liters will be:
Volume of solution = Volume of solution (in mL) / 1000
Volume of solution = 1 mL / 1000
Volume of solution = 0.001 L
Finally, we can calculate the molarity of the solution using the number of moles of NaClO and the volume of the solution in liters:
Molarity = Number of moles of NaClO / Volume of solution (in L)
Molarity = 0.000706 mol / 0.001 L
Molarity = 0.706 M
Therefore, the molarity of the solution is 0.706 M.