Caint Tlgtres. Calculation Zone B. Standard 0.01 M Na S20, Solution 8. Average m
ID: 229644 • Letter: C
Question
Caint Tlgtres. Calculation Zone B. Standard 0.01 M Na S20, Solution 8. Average molar concentration 0.135 Data Analysis 1. of Na2S203 (mol/L) C. Preparation of Liquid Bleach for Analysis 1. Volume of original bleach Data Analysis 2. titrated (mL) 2. Mass of bleach titrated (g) (assume the density of the liquid bleach is 1.084 g/mL) E. Analysis of Bleach Samples 1. Buret reading, initial (mL) 2. Buret reading, final (mL) 3. Volume of Na S,O Data Analysis 3. 3.85 37.60 added (mL) Data Analysis Data Analysis 4 1. Moles of Na S203 added (mol) Show calculation. 2. Moles of CIO reacted (mol) Equations 29.4 and 29.3 Show calculation. Moles of available chlorine (mol) Equation 29.2. Show calculation. Mass of available chlorine (g) Show calculation Mass percent available chlorine (liquid) (%) Show calculation. 3. Data Analysis 5. 4. 5.Explanation / Answer
Molarity of Na2S2O3 = 0.135 g/mol = 0.135 M
C. Preparation of liquid bleach for analysis.
Vol. of original bleach titrated = 2.50 mL
Density of the liquid bleach = 1.084g/mL
That is, 1 mL of the liquid bleach has bleach of mass 1.084g
So, 2.50 mL of the liquid bleach will have mass
= 1.084 g x 2.50
= 2.71 g
Therefore, mass of bleach titrated = 2.71 g
E. Analysis of bleach samples
Volume of Na2S2O3 added = final burette reading – initial burette reading
V = 37.60 mL – 3.85 mL
V = 33.75 mL
Therefore, volume of Na2S2O3 added = 33.75 mL
Data Analysis 4:
No. of moles of Na2S2O3 added
Volume of Na2S2O3 added, V = 33.75 mL = 33.75 x 10-3 L
Molarity of Na2S2O3, M= 0.135mol/L
No. of moles can be calculated using
n = M x V
where n = no. of moles
M = molarity
V = volume in L
Therefore,
n = 0.135 mol/L x 33.75 x 10-3 L
n = 4.5563 x 10-3 mol
Therefore, no. of moles of Na2S2O3 added = 4.5563 x 10-3 mol
Moles of ClO- reacted
Volume of liquid bleach = 2.50 mL
Molarity of Na2S2O3 = 0.135mol/L
Volume of Na2S2O3 added = 33.75 mL
(MV)liq. bleach = (MV)thio
Mliq. bleach = (MV)thio/ Vliq.bleach
Mliq. bleach = (0.135 mol/L x 33.75 mL)/ 2.50 mL
Mliq. bleach = 1.8225 mol/L
Therefore, no. of moles in the liquid bleach will be
n = Mliq. bleach x Vol. taken (in L)
n = 1.8225 mol/L x (2.50 x 10-3) L
n = 4.5563 x 10-3 mol
Therefore, moles of ClO- reacted = 4.5563 x 10-3 mol
Data Analysis 5:
Mass and no. of moles of available chlorine
Molar mass of NaClO = 74.44 g/mol
No. of moles is given by
n = m/MM
where m = mass
MM = molar mass
Therefore,
m = n x MM
m = (4.5563 x 10-3)mol x 74.44 g/mol
m = 0.3392 g
That is the liquid bleach have 0.3392g of NaClO.
74.44 g of NaClO has 35.5 g of Cl
Therefore, 0.3392 g of NaClO will have Cl of mass
= (35.5g/74.44) x 0.3392
= 0.1618 g of Cl
Therefore, mass of available chlorine = 0.1618 g
No. of moles of available chlorine
n = mass/atomic mass
n = 0.1618g/35.5 g mol-1
n = 4.5577 x 10-3 mol
No. of moles of available chlorine = 4.5577 x 10-3 mol
Mass % of available chlorine
Mass % of Chlorine is calculated using formula
Mass % = (mass of available chlorine/mass of liquid bleach) x 100
Mass % = (0.1618g/2.71g) x 100
Mass % = 5.97%
Therefore, mass % of available chlorine is 5.97%