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48. Determine the molarity of each of the following solutions:

(a) 1.457 mol KCl in 1.500 L of solution

(b) 0.515 g of H2SO4 in 1.00 L of solution

(c) 20.54 g of Al(NO3)3 in 1575 mL of solution

(d) 2.76 kg of CuSO4∙5±á2O in 1.45 L of solution

(e) 0.005653 mol of Br2 in 10.00 mL of solution

(f) 0.000889 g of glycine, C2H5NO2, in 1.05 mL of solution

Short Answer

Expert verified

a) 0.971 M

b) 0.0052 M

c) 0.0603 M

d) 7.644 M

e)0.05653 M

f) 0.0112 M

Step by step solution

01

Subpart a)

1.457 mol KCl in 1.500 L of solution

Mole of KCl = 1.457

Molarity = Moles of solute / Liters of solution.

Molarity = 1.457 mol/ 1.5 L = 0.971 M

02

Subpart b)

0.515 g of H2SO4 in 1.00 L of solution

Mass of H2SO4 = 0.515 g

Molecular weight of H2SO4 = 2x 1 + 32 + 4 x 16 = 98 g/mol

Moles of H2SO4 = 0.515 g / 98 g/mol = 0.0052 Moles

Molarity = Moles of solute/ Liters of solution

= 0.0052 Moles/ 1 L

= 0.0052 M

03

Subpart c)

20.54 g of Al(NO3)3 in 1575 mL of solution

1575 mL = 1.575 L ( 1 L = 1000 mL)

Mass of Al(NO3)3 = 20.54 g

Molecular weight of Al(NO3)3 = 30 + 3 x 14 + 9 x 16 = 30 + 42+ 144 = 216 g/mol

Moles of Al(NO3)3 = 20.54 g/ 216 g/mol = 0.0950 mole

Molarity = 0.0950 mole/ 1.575 L = 0.0603 M

04

Subpart d)

2.76 kg of CuSO4∙5±á2O in 1.45 L of solution

Mass ofCuSO4∙5±á2O = 2.76 kg = 2760 g (1kg = 1000 g)

Molecular weight of CuSO4∙5±á2O = 63 + 32 + 9 x 16 + 10 x 1 = 95 + 154 = 249 g/mol

Moles of CuSO4∙5±á2O = 2760 g / 249 g/mol = 11.084 mole

Molarity of CuSO4∙5±á2O = 11.084 mole/ 1.45 L

= 7.644 M

05

Subpart e)

0.005653 mol of Br2 in 10.00 mL of solution

10 mL = 0.01L ( 1L = 1000 mL )

Molarity of Br2 = Moles of solute / Liters of solution

Molarity of Br2 = 0.005653 mol/ 0.1 L

= 0.05653 M

06

Subpart f)

0.000889 g of glycine, C2H5NO2, in 1.05 mL of solution

1.05 mL = 0.00105 L ( 1 L = 1000 mL )

Molecular weight of C2H5NO2 = 2x 12 + 5 x 1+ 14 + 2 x 16 = 29 + 14 + 32 = 75 g/mol

Moles of C2H5NO2 = 0.000889 g / 75 g/mol = 1.185 x 10-5 mole

Molarity = 1.185 x 10-5 mole/ 0.00105 L

= 0.0112 M

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