Answered

1. In preparing Solution 3 for the kinetics rate runs, a student weighed our 0.2653 g of dye (molecular weight 879.9 g/mol. This was transferred to a 100 mL volumetric flask and water was add to make 100.00 mL of solution 1. A 10.00 mL aliquot of solution 1 was diluted to 100.00 mL to make solution 2 and a 5.00 mL aliquot of solution 2 was then diluted to 100.00 mL to make solution 3. Calculate the concentration of solution 3 in weight/volume percent (w/v%), parts per million ( ppm), and molarity (M). Report the correct number of significant figures.

Answer :

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Answer:

Explanation:

Weight of the dye in  solution 1 = 0.2653 g in 100 mL

Weight of the dye in  solution 2 = 0.02653 g in 100 mL

Weight of the dye in  solution 1 = 0.0013265 g in 100 mL

The percentage concentration can be calculated by using the formula :

percentage concentration = (weight of the substance / volume of the solution) × 100

For Solution (1)

percentage concentration = (0.2653 /100) × 100

percentage concentration = 0.2653%

percentage concentration = 0.267 % (to  significant figure)

For solution (2)

percentage concentration = (0.02653 /100) × 100

percentage concentration = 0.02653%

percentage concentration = 0.0267 % (to  significant figure)

For the final percentage concentration for Solution (3)

percentage concentration = (0.0013265 /100) × 100

percentage concentration = 0.0013265%

percentage concentration =  0.0013 % (to  significant figure)

The mg/L for solution (1) is:

= 265.3 mg/0.1 L

= 2653 mg/L

The mg/L for solution (2) is:

= 26.53 mg/0.1 L

= 265.3 mg/L

The mg/L for solution (3) is:

= 1.3265  mg/0.1 L

= 13.265  mg/L

Molarity = (given weight/molecular weight) × (volume of the solution in L / 1.0 L)

For solution (1):

Molarity  = (0.2653 / 879.9)  × (0.1 L /1 L)

               = 3.128×10⁻⁵ M

For solution (2)

Molarity = (0.02653 / 879.9) × (0.1 L / 1 L)

             = 3.015×10⁻⁶ M

For solution (3)

Molarity =  (0.0013265 / 879.9) × (0.1 L / 1 L)

= 1.508×10⁻⁷M

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