Question

In: Chemistry

Determine the w/w% of chloride in unknown. 6.4923g of AgNO3 is dissolved in 500mL of water...

Determine the w/w% of chloride in unknown.

6.4923g of AgNO3 is dissolved in 500mL of water producing a 0.076438M solution.

1.4838g of CaCl2 is dissolved into 250mL of water producing a 5.3479x10^(-2)M solution.

10.00mL pipette is calibrated to 9.927mL

25.00mL pipette is calibrated to 25.163mL

Each of th five flasks contains 25.163mL of the CaCl2 solution

For the unknown each of the five flasks contains 25.163mL

Titrations

CaCl2 STD Starting vol End vol
1 .26 37.58
2 1.44 36.71
3 .87 36.14
4 1.05 37.80
5 1.22 36.51

UNKNOWN

Unknown starting vol end vol
1 1.20 24.14
2 .94 23.19
3 .65 23.93
4 .61 22.84
5 .84 23.03

The first titrations of each round were rough titrations and should not be included

Solutions

Expert Solution

Titration : Unknown

Run 2. moles of AgNO3 used = 0.076438M x (23.19 - 0.94) ml = 1.70 mmol

moles of CaCl2 reacted = 1.70/2 = 0.85 mmol

concentration [CaCl2] in unknown = 0.85 mmol/25.163 ml = 0.034 M

CaCl2 = 0.034 x 110.98 = 3.77 g

1 CaCl2 has 2 Cl in it

[Cl] in unknown = 2 x 0.034 mol/L x 35 g/mol = 2.38 g

w/w% Cl = (2.38/3.77) x 100 = 63.13%

Similarly for other runs calculation is done,

Run 3. moles of AgNO3 used = 0.076438M x (23.93 - 0.65) ml = 1.78 mmol

moles of CaCl2 reacted = 1.78/2 = 0.89 mmol

concentration [CaCl2] in unknown = 0.89 mmol/25.163 ml = 0.035 M

CaCl2 = 0.035 x 110.98 = 3.88 g

1 CaCl2 has 2 Cl in it

[Cl] = 2 x 0.035 mol/L x 35 g/mol = 2.45 g

w/w% Cl = (2.45/3.88) x 100 = 63.14%

Run 4. moles of AgNO3 used = 0.076438M x (22.84 - 0.61) ml = 1.70 mmol

moles of CaCl2 reacted = 1.7/2 = 0.85 mmol

concentration [CaCl2] in unknown = 0.85 mmol/25.163 ml = 0.034 M

CaCl2 = 0.034 x 110.98 = 3.77 g

[Cl] = 2 x 0.034 mol/L x 35 g/mol = 2.38 g

w/w% Cl = (2.38/3.77) x 100 = 63.13%

Run 5. moles of AgNO3 used = 0.076438M x (23.03 - 0.84) ml = 1.70 mmol

moles of CaCl2 reacted = 1.70/2 = 0.85 mmol

concentration [CaCl2] in unknown = 0.85 mmol/25.163 ml = 0.034 M

CaCl2 = 0.034 x 110.98 = 3.77 g

[Cl] = 2 x 0.034 mol/L x 35 g/mol = 2.36 g

w/w% Cl = (2.36/3.77) x 100 = 62.60%


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