Question

Determine the molarity for a solution of na3po4 prepared by diluted 10.0 ml of a 0.33 m solution of na3po4 to 1.00 l. 2. if 5.00 ml of the 1.0 l solution prepared in question 1 was diluted to 50.0 ml, what would be the concentration of na3po4 in this new solution? 3. the absorbance of a 0.100 mm solution of a

Answer 1

Answer i) The molarity for the solution of $Na_{3} PO_{4}$ prepared after dilution from 10 mL of 0.33 M solution to 1 L solution is 3.3 X $10^{-3}$,

$m _{1}V_{1}=m_{2} V_{2}$ by using this formula. 

So, Molarity of unknown solution = (10 X 0.33) / 1000 = 3.3 X $10^{-3}$.

Answer ii) When you dilute the solution prepared in question (1 ) to 50.0 mL from 5.00 mL of 1L solution then, you should take an aliquot of 5 mL and then dilute upto 50 mL that will be the result of another 10 times dilution which will result in the new concentration as 0.00033 M for $Na_{3} PO_{4}$ solution.

Answer iii)
Question iii is incomplete. Complete question is attached below;
Solution:
Given: A = 0.982, b = 1 cm, c = 0.0001 M

we know that, A = ebc
Therefore, ∈ = A/bc = 0.982/(1 X 0.0001) = 9820 M-1 cm-1.

Thus, molar absorptivity of solution is 9820 M-1 cm-1.