Question

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.

Answer 1

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 = $\frac{19.5 g}{58.5 g/mol}= 0.3333 mol$

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

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

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

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

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

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

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)

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

The solutions in order of decreasing concentration:

(IV) > (II) > (III) > (I)  = (V)

Answer 2

Answer:

See the answers below

Explanation:

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

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

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

4)  Number 1 and 5 have the same concentration (1.5M)

MW of NaCl = 23 + 36 = 59 g

For 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 number 4,

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

For number 5

Molarity (M) = 0.450/0.3 = 1.5 M