Question

question 2 (1 point)
how many ml of a 5.00 ppm solution are needed to have 1.00 g ?
the concentration will always be the conversion factor. like %m/m, ppm and related
concentrations are all in g/ml.
5.00 ppm means 5.00 g in a million ml, or 5.00 g/$10^6$ ml, or 5.00 g/ml x $10^6$.
1 ml
(2 g)(---------) = 2.00 x 3 ml
4 g
a. 0.00 b. 1.00 c. 2.00 d. 3.00 e. 4.00 f. 5.00
g. 6.00 h. 8.00 i. 12.00 j. 2000 k. $10^2$ l. $10^5$
m. $10^6$ n. $10^9$ o. $10^{12}$ p. $10^{-2}$ q. $10^{-3}$ r. $10^{-6}$
s. $10^{-9}$ t. $10^{-12}$ u. $10^8$ v. $10^{-1}$ w. $10^1$ x. 30.0
y. 0.100 z. 0.200 aa. 20.0 bb. 40.0
cc. 1.33 dd. 0.267

Explanation:

Step1: Understand ppm definition

5.00 ppm means \( \frac{5.00\ \text{g}}{10^6\ \text{mL}} \), so the conversion factor is \( \frac{10^6\ \text{mL}}{5.00\ \text{g}} \).

Step2: Set up the calculation

We need to find volume (mL) for 1.00 g. The formula is \( \text{Volume} = \text{Mass} \times \frac{1}{\text{Concentration (g/mL)}} \). Substituting values: \( (1.00\ \text{g}) \times \frac{10^6\ \text{mL}}{5.00\ \text{g}} \).

Step3: Calculate the volume

Simplify: \( \frac{1.00 \times 10^6}{5.00} = 2.00 \times 10^5\ \text{mL} \). Now match to the blanks:

• Blank 2: Mass is 1.00 g, so option B.
• Blank 4: The numerator of the conversion factor's inverse (for g) is 5.00? Wait, no—wait, the conversion factor is \( \frac{10^6\ \text{mL}}{5.00\ \text{g}} \), so the denominator in the fraction (blank 4) is 5.00 g? Wait, no, the structure is \( (\text{mass g}) \times (\frac{\text{mL}}{\text{g}}) \). So the fraction is \( \frac{10^6\ \text{mL}}{5.00\ \text{g}} \), so blank 4 (the g in denominator) is 5.00? Wait, no, the problem's equation: \( (\_2\_\ \text{g})(\frac{\_1\_\ \text{mL}}{\_4\_\ \text{g}}) = 2.00 \times \_3\_\ \text{mL} \). So:
• Blank 2: 1.00 g (B)
• Blank 4: 5.00 g (F)
• Blank 3: \( 10^5 \) (L)
• Wait, let's re-express the calculation: \( 1.00\ \text{g} \times \frac{10^6\ \text{mL}}{5.00\ \text{g}} = \frac{1.00 \times 10^6}{5.00}\ \text{mL} = 2.00 \times 10^5\ \text{mL} \). So:
• Blank 2: 1.00 (B)
• Blank 4: 5.00 (F)
• Blank 3: \( 10^5 \) (L)
• Wait, the first blank (1) is mL, the fraction is \( \frac{10^6\ \text{mL}}{5.00\ \text{g}} \), so the numerator of the fraction (blank 1's mL) is \( 10^6 \)? No, wait the equation is \( (\_2\_\ \text{g})(\frac{\_1\_\ \text{mL}}{\_4\_\ \text{g}}) = 2.00 \times \_3\_\ \text{mL} \). Let's plug numbers: \( (1.00\ \text{g})(\frac{10^6\ \text{mL}}{5.00\ \text{g}}) = 2.00 \times 10^5\ \text{mL} \). So:
• Blank 2: 1.00 (B)
• Blank 4: 5.00 (F)
• Blank 3: \( 10^5 \) (L)
• Wait, but the problem's equation has "2.00 x _3_ mL", and our result is \( 2.00 \times 10^5 \), so _3_ is \( 10^5 \) (L).
• Blank 1: The mL in numerator is \( 10^6 \)? No, the fraction is \( \frac{10^6\ \text{mL}}{5.00\ \text{g}} \), so blank 1 (mL) is \( 10^6 \)? Wait, no, the equation is \( (1.00\ \text{g})(\frac{10^6\ \text{mL}}{5.00\ \text{g}}) = 2.00 \times 10^5\ \text{mL} \). So:
• Blank 2: B (1.00)
• Blank 4: F (5.00)
• Blank 3: L (10^5)
• Wait, maybe the first blank (1) is \( 10^6 \)? No, the fraction is \( \frac{10^6\ \text{mL}}{5.00\ \text{g}} \), so blank 1 (mL) is \( 10^6 \)? But the calculation gives \( 2.00 \times 10^5 \), so:
• \( 1.00 \times \frac{10^6}{5.00} = 2.00 \times 10^5 \), so the exponents: \( 10^6 / 5 = 2 \times 10^5 \), yes. So:
• Blank 2: B (1.00)
• Blank 4: F (5.00)
• Blank 3: L (10^5)
• Wait, the problem's options for blank 3: L is \( 10^5 \), yes.

Answer:

Blank 2: B. 1.00
Blank 4: F. 5.00
Blank 3: L. \( 10^5 \)

(Note: Assuming the structure, the key blanks are 2 (B), 4 (F), 3 (L). The calculation confirms \( 1.00\ \text{g} \times \frac{10^6\ \text{mL}}{5.00\ \text{g}} = 2.00 \times 10^5\ \text{mL} \), so blank 3 is \( 10^5 \) (L), blank 2 is 1.00 (B), blank 4 is 5.00 (F).)