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. $10^6$ i 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: Recall ppm definition

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

Step2: Set up the calculation

We need to find volume (\(V\)) when mass (\(m = 1.00\ \text{g}\)) is known. Using \(V=m\times\text{conversion factor}\), substitute \(m = 1.00\ \text{g}\) and conversion factor \(\frac{10^6\ \text{mL}}{5.00\ \text{g}}\).

$$V = 1.00\ \text{g}\times\frac{10^6\ \text{mL}}{5.00\ \text{g}}$$

Step3: Perform the calculation

Simplify the expression: \(\frac{1.00\times10^6}{5.00}=2.00\times10^5\ \text{mL}\). Now, let's match the blanks:

• Blank 2: The mass is \(1.00\ \text{g}\), so option B.
• Blank 4: The mass in the conversion factor's denominator is \(5.00\ \text{g}\), so option F.
• Blank 3: The exponent for \(10\) is \(5\), so option L (\(10^5\)).
• Blank 1: The numerator of the conversion factor's volume part is \(10^6\), but in the given formula structure, let's re - check the formula: \((\underline{2}\ \text{g})(\frac{\underline{1}\ \text{mL}}{\underline{4}\ \text{g}})=2.00\times\underline{3}\ \text{mL}\). Wait, actually, from \(V = m\times\frac{10^6\ \text{mL}}{5.00\ \text{g}}\), we can rewrite it as \(m\times\frac{10^6/5.00\ \text{mL}}{\text{g}}\). \(10^6/5.00 = 2.00\times10^5\), so:
• Blank 2: \(1.00\ \text{g}\) (option B)
• Blank 4: \(5.00\ \text{g}\) (option F)
• Blank 1: \(10^6\ \text{mL}\) (but in the formula's numerator of the fraction, it's \(10^6\) related? Wait, no, let's use the formula given: \((\underline{2}\ \text{g})(\frac{\underline{1}\ \text{mL}}{\underline{4}\ \text{g}})=2.00\times\underline{3}\ \text{mL}\). So:
• \(\underline{2}\) is the mass, which is \(1.00\ \text{g}\) (B)
• \(\underline{4}\) is the mass in the denominator, which is \(5.00\ \text{g}\) (F)
• \(\underline{1}\) is related to the volume part, and when we calculate \(1.00\times\frac{10^6}{5.00}=2.00\times10^5\), so \(\underline{3}\) is \(10^5\) (L)
• \(\underline{1}\) in the numerator of the fraction: from \(\frac{10^6\ \text{mL}}{5.00\ \text{g}}\), we can think of it as \(\frac{10^6/5.00\ \text{mL}}{\text{g}}\), and \(10^6/5.00 = 2.00\times10^5\), but in the formula structure, the first blank (numerator of the fraction) should be \(10^6\) (but in the options, M is \(10^6\)? Wait, no, let's re - express the formula correctly. The correct calculation gives us that for the formula \((m\ \text{g})(\frac{V_{per\ g}\ \text{mL}}{\text{g}})=V\ \text{mL}\), where \(V_{per\ g}=\frac{10^6}{5.00}\ \text{mL/g}=2.00\times10^5\ \text{mL/g}\). So in the given formula \((\underline{2}\ \text{g})(\frac{\underline{1}\ \text{mL}}{\underline{4}\ \text{g}})=2.00\times\underline{3}\ \text{mL}\), \(\underline{1}\) should be \(10^6\) (M), \(\underline{2}\) is \(1.00\) (B), \(\underline{4}\) is \(5.00\) (F), and \(\underline{3}\) is \(10^5\) (L).

Answer:

• Blank 2: B. 1.00
• Blank 4: F. 5.00
• Blank 3: L. \(10^5\)
• Blank 1: M. \(10^6\) (assuming the formula structure is \((1.00\ \text{g})(\frac{10^6\ \text{mL}}{5.00\ \text{g}})=2.00\times10^5\ \text{mL}\))