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

Question

7 days ago

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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.

Chemistry

2 answers:

lions [1K] · Answer 1 · 2026-02-26T16:57:42+03:00

Answer:

The order of the solutions based on decreasing concentration:

(IV) > (II) > (III) > (I) = (V)

Explanation:

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

Given concentration of NaCl is [NaCl]= 1.50 M

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

Given concentration of NaCl is [NaCl]= 3.00 M

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

Calculated moles of NaCl = $\frac{19.5 g}{58.5 g/mol}= 0.3333 mol$

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

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

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

Calculated moles of NaCl = $\frac{19.5 g}{58.5 g/mol}= 0.3333 mol$

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

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

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

Calculated moles of NaCl = 0.450 mol

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

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

The order of the solutions based on decreasing concentration:

(IV) > (II) > (III) > (I) = (V)

Anarel [852] · Answer 2 · 2026-02-26T16:59:50+03:00

Answer:

Please review the following responses

Explanation:

1) A solution of 100. mL contains 19.5 g of NaCl (3.3M)

2) 100. mL of NaCl solution at 3.00 M (3 M)

3) A solution of 150. mL holds 19.5 g of NaCl (2.2 M)

4) The concentrations of beakers 1 and 5 are identical (1.5M)

Molar mass of NaCl = 23 + 36 = 59 g

For beaker number 3:

59 g -------------- 1 mol

19.5 g ------------- x

x = 19.5 x 1/59 = 0.33 mol

Molarity (M) = 0.33 mol/0.150 l = 2.2 M

For beaker number 4:

Molarity (M) = 0.33mol/0.10 l = 3.3 M

For beaker number 5:

Molarity (M) = 0.450/0.3 = 1.5 M