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Calculation of Molarities and Volumes of Solutions, Exercises of Chemistry

University of Saint Francis (USF)Chemistry

Calculations for determining the molarities and volumes of various solutions based on given volumes and molarities. It includes conversions between moles, liters, and milliliters, as well as the calculation of equivalent volumes and molarities for different reactions.

Typology: Exercises

2021/2022

Uploaded on 09/27/2022

anjushri
anjushri 🇺🇸

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1. (a) mol of FeCl3 = 11.4 g FeCl x = 0.07028 mol FeCl3
1 mol FeCl3
162.2 g FeCl3
mL of soln = 0.07028 mol FeCl3x = 68.6 mL
1000 mL
1.025 mol FeCl3
(b) mol FeCl3 = 555 mL x = 0.569 mol FeCl3
1000 mL
1.025 mol FeCl3
(c) mol FeCl3 = 125 mL x = 0.1281 mol FeCl3
1000 mL
1.025 mol FeCl3
mol Cl- = 0.1281 mol FeCl3 x = 0.384 mol Cl-
3 mol Cl-
1 mol FeCl3
(d) mol FeCl3 = 1.65 L x = 1.691 mol FeCl3
1.025 mol FeCl3
1 L
grams FeCl3 = 1.691 mol FeCl3 x = 274 g FeCl3
1 mol FeCl3
162.2 g FeCl3
(e) MoVo = MdVd Vd = (25.0 + 125.0) mL = 150.0 mL
Md = = 0.171 M
1.025 M x 25.0 mL
150.0 mL
(f) Vo = = 45.1 mL
0.0925 M x 500.0 mL
1.025 M
(g) Mresulting solution = ntotal
Vtotal(L)
ntotal = nsolution 1 + nsolution 2
= 75.0 mL x + 50.0 mL x
1.025 mol
1000 mL 1000 mL
1.875 mol
= (0.076875 + 0.09375) mol = 0.17063 mol
Vtotal = 75.0 mL + 50.0 mL = 125.0 mL = 0.1250 L
Mresulting solution = = 1.365 M
0.17063 mol
0.1250 L
CHEQ 1094 SOLUTIONS PROBLEM SET: ANSWERS
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  1. (a) mol of FeCl3 = 11.4 g FeCl x = 0.07028 mol FeCl (^3)

1 mol FeCl 3 162.2 g FeCl

mL of soln = 0.07028 mol FeCl 3 x = 68.6 mL 1000 mL 1.025 mol FeCl 3

(b) mol FeCl3 = 555 mL x (^) 1000 mL = 0.569 mol FeCl 3

1.025 mol FeCl 3

(c) mol FeCl3 = 125 mL x (^) 1000 mL = 0.1281 mol FeCl 3

1.025 mol FeCl 3

mol Cl-^ = 0.1281 mol FeCl 3 x = 0.384 mol Cl-

3 mol Cl- 1 mol FeCl 3

(d) mol FeCl3 = 1.65 L x = 1.691 mol FeCl 3

1.025 mol FeCl 1 L

grams FeCl 3 = 1.691 mol FeCl 3 x (^) 1 mol FeCl = 274 g FeCl (^3) 3

162.2 g FeCl 3

(e) MoVo = M (^) dVd Vd = (25.0 + 125.0) mL = 150.0 mL

M (^) d = = 0.171 M 1.025 M x 25.0 mL 150.0 mL

(f) V (^) o = = 45.1 mL 0.0925 M x 500.0 mL 1.025 M

(g) M (^) resulting solution =

n total Vtotal(L)

n total = n solution 1 + n solution 2

= 75.0 mL x + 50.0 mL x

1.025 mol 1000 mL 1000 mL

1.875 mol

= (0.076875 + 0.09375) mol = 0.17063 mol

Vtotal = 75.0 mL + 50.0 mL = 125.0 mL = 0.1250 L

Mresulting solution = = 1.365 M

0.17063 mol 0.1250 L

  1. (a) mol Fe(OH) 3 = 1.38 g Fe(OH)3 x = 0.01291 mol Fe(OH) 3

1 mol Fe(OH) 106.9 g Fe(OH) 3

mol FeCl3 = mol Fe(OH)3 = 0.01291 mol

mL FeCl 3 = 0.01291 mol FeCl 3 x = 109 mL FeCl (^3)

1000 mL FeCl 3 0.1189 mol FeCl 3

(b) mol FeCl3 = 25.0 mL x (^) 1000 mL FeCl = 0.002973 mol FeCl (^3) 3

0.1189 mol FeCl 3

mol KOH = 0.002973 mol FeCl3 x = 0.008919 mol KOH

3 mol KOH 1 mol FeCl (^3)

mL KOH = 0.008919 mol KOH x = 43.5 mL KOH 1000 mL KOH 0.205 mol KOH

(c) mol KOH = 0.184 g KOH x (^) 56.1 g KOH = 0.003280 mol KOH 1 mol KOH

mol FeCl 3 = 0.003280 mol KOH x (^) 3 mol KOH = 0.001093 mol FeCl

1 mol FeCl 3

mL FeCl3 = 0.001093 mol FeCl 3 x = 9.19 ml FeCl (^3)

1000 mL FeCl 3 0.1189 mol FeCl 3

(d) g of KOH = 0.265 g imp KOH x (^) 100 g imp KOH = 0.2253 g KOH

85.0 g KOH

mol KOH = 0.2253 g KOH x (^) 56.1 g KOH = 0.004016 mol KOH 1 mol KOH

mol FeCl 3 = 0.004016 mol KOH x (^) 3 mol KOH = 0.001339 mol FeCl

1 mol FeCl 3

mL FeCl3 = 0.001339 mol FeCl x = 11.3 mL FeCl (^3)

1000 mL FeCl 3 0.1189 mol FeCl 3