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The following five beakers, each containing a solution of sodium chloride (NaCl, also known as table salt), were found on a lab shelf:

Beaker Contents
1) 200. mL of 1.50 M NaCl solution
2) 100. mL of 3.00 M NaCl solution
3) 150. mL of solution containing 19.5 g of NaCl
4) 100. mL of solution containing 19.5 g of NaCl
5) 300. mL of solution containing 0.450 mol NaCl

Arrange the solutions in order of decreasing concentration.

Respuesta :

Answer:

The solutions in order of decreasing concentration:

(IV) > (II) > (III) > (I)  = (V)

Explanation:

1) 200 mL of 1.50 M NaCl solution  - (I)

Concentration of NaCl is given , [NaCl]= 1.50 M

2) 100 mL of 3.00 M NaCl solution  - (II)

Concentration of NaCl is given , [NaCl]= 3.00 M

3) 150 mL of solution containing 19.5 g of NaCl  - (III)

Moles of NaCl = [tex]\frac{19.5 g}{58.5 g/mol}= 0.3333 mol[/tex]

Volume of solution = 150 mL = 0.150 (1L = 1000 mL)

[tex][NaCl]=\frac{0.3333 mol}{0.150 L}=2.222 M[/tex]

4) 100 mL of solution containing 19.5 g of NaCl  - (IV)

Moles of NaCl = [tex]\frac{19.5 g}{58.5 g/mol}= 0.3333 mol[/tex]

Volume of solution = 100 mL = 0.100 (1L = 1000 mL)

[tex][NaCl]=\frac{0.3333 mol}{0.100 L}=3.333 M[/tex]

5) 300 mL of solution containing 0.450 mol NaCl - (V)

Moles of NaCl = 0.450 mol

Volume of solution = 300 mL = 0.300 (1L = 1000 mL)

[tex][NaCl]=\frac{0.450 mol}{0.300 L}=1.50 M[/tex]

The solutions in order of decreasing concentration:

(IV) > (II) > (III) > (I)  = (V)

dsdrajlin

We have 5 beakers with different solutions of NaCl. The order of the solutions in order of decreasing concentration is: 4, 2, 3, 1 = 5.

We have 5 beakers with different solutions of NaCl. To compare their concentrations, we will calculate the molarity of each solution.

1) 200. mL of 1.50 M NaCl solution

The molarity of the solution is 1.50 M.

2) 100. mL of 3.00 M NaCl solution

The molarity of the solution is 3.00 M.

3) 150. mL of solution containing 19.5 g of NaCl

We will use the following expression.

[tex]M = \frac{mass\ solute }{molar\ mass\ solute \times liters\ solution} = \frac{19.5g}{58.44g/mol \times 0.150L} = 2.22 M[/tex]

4) 100. mL of solution containing 19.5 g of NaCl

We will use the following expression.

[tex]M = \frac{mass\ solute }{molar\ mass\ solute \times liters\ solution} = \frac{19.5g}{58.44g/mol \times 0.100L} = 3.34 M[/tex]

5) 300. mL of solution containing 0.450 mol NaCl

We will use the following expression.

[tex]M = \frac{moles\ solute }{liters\ solution } = \frac{0.450mol}{0.300L} = 1.50M[/tex]

The order of the solutions in order of decreasing concentration is:

4, 2, 3, 1  = 5

We have 5 beakers with different solutions of NaCl. The order of the solutions in order of decreasing concentration is: 4, 2, 3, 1 = 5.

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