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What is the concentration of [tex]$F_2( g )$[/tex], in parts per billion, in a solution that contains [tex]$4.8 \times 10^{-8} g$[/tex] of [tex][tex]$F_2( g )$[/tex][/tex] dissolved in [tex]$9.6 \times 10^{-3} g$[/tex] of [tex]$H_2 O ( l )$[/tex]?

A. [tex][tex]$5 ppb$[/tex][/tex]
B. [tex]$0.5 ppb$[/tex]
C. [tex]$5.0 \times 10^2 ppb$[/tex]
D. [tex][tex]$5.0 \times 10^3 ppb$[/tex][/tex]

Sagot :

GinnyAnswer
To determine the concentration of [tex]\( F_2(g) \)[/tex] in parts per billion (ppb) in a solution, we follow these steps:

1. Identify the given masses:
- Mass of [tex]\( F_2(g) \)[/tex]: [tex]\( 4.8 \times 10^{-8} \)[/tex] grams
- Mass of [tex]\( H_2O(l) \)[/tex]: [tex]\( 9.6 \times 10^{-3} \)[/tex] grams

2. Understand parts per billion (ppb):
- Parts per billion is a ratio where 1 ppb means 1 part of the substance per [tex]\( 10^9 \)[/tex] parts of the solution.

3. Set up the ratio:
[tex]\[ \text{Concentration (ppb)} = \left( \frac{\text{Mass of } F_2(g)}{\text{Mass of solution}}\right) \times 10^9 \][/tex]

4. Substitute the given values into the ratio:
[tex]\[ \text{Concentration (ppb)} = \left( \frac{4.8 \times 10^{-8} \text{ g } F_2(g)}{9.6 \times 10^{-3} \text{ g } H_2O(l)} \right) \times 10^9 \][/tex]

5. Simplify the fraction:
[tex]\[ \frac{4.8 \times 10^{-8}}{9.6 \times 10^{-3}} = \frac{4.8}{9.6} \times 10^{-8 + 3} = 0.5 \times 10^{-5} \][/tex]

6. Calculate the concentration:
[tex]\[ \text{Concentration (ppb)} = 0.5 \times 10^{-5} \times 10^9 = 0.5 \times 10^{4} = 5000 \text{ ppb} \][/tex]

This result matches with option D.

Therefore, the concentration of [tex]\( F_2(g) \)[/tex] in the solution is [tex]\( \boxed{5.0 \times 10^3 \text{ ppb}} \)[/tex].
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