Question

1.00 mL of a 3.55 × 10–4 M solution of oleic acid is diluted with 9.00 mL of petroleum ether, forming solution A. Then 2.00 mL of solution A is diluted with 8.00 mL of petroleum ether, forming solution B. What is the concentration of solution B?

Answer 1

Answer:

Concentration of solution B is $7.10\times 10^{-6}M$

Explanation:

Solution A: Total volume of solution A = (9.00+1.00) mL = 10.00 mL

According to law of dilution, $C_{1}V_{1}=C_{2}V_{2}$

where $C_{1}$ and $C_{2}$ are initial and final concentration respectively. $V_{1}$ and $V_{2}$ are initial and final volume respectively.

Here $C_{1}$ = $3.55\times 10^{-4}M$, $V_{1}$ = 1.00 mL, $V_{2}$ = 10.00 mL

So, $C_{2}=\frac{C_{1}V_{1}}{V_{2}}$ = $\frac{3.55\times 10^{-4}M\times 1.00mL}{10.00mL}=3.55\times 10^{-5}M$

So, concentration of solution A = $3.55\times 10^{-5}M$

Solution B: Total volume of solution B = (2.00+8.00) mL = 10.00 mL

Similarly as above, $C_{2}=\frac{C_{1}V_{1}}{V_{2}}$ = $\frac{3.55\times 10^{-5}M\times 2.00mL}{10.00mL}=7.10\times 10^{-6}M$

So, concentration of solution B = $7.10\times 10^{-6}M$