Question

title = q7a10 the temperature of a chemistry lab you are working in is 25.000 °c. this is important to know, because at this temperature the density of water is 0.99704 g/cm³ (the density of water changes at different temperatures as we will discuss later), and you need this conversion factor for this problem. you use a device called a graduated cylinder flask to provide 100. ml of water. you now dissolve 58.44 grams of sodium chloride in your 100. ml of water. what is the total mass of the solution you made, reported to the proper number of significant figures?
158.144 g
158.14 g
158.1 g
158 g
1.6 × 10² g

Explanation:

Step1: Calculate mass of water

Density formula: $
ho = \frac{m}{V}$, so $m =
ho \times V$.
Given $
ho = 0.99704\ \text{g/cm}^3$, $V = 100.\ \text{mL} = 100.\ \text{cm}^3$ (since $1\ \text{mL} = 1\ \text{cm}^3$).
$m_{\text{water}} = 0.99704\ \text{g/cm}^3 \times 100.\ \text{cm}^3 = 99.704\ \text{g}$.

Step2: Calculate total mass of solution

Total mass = mass of water + mass of NaCl.
$m_{\text{total}} = 99.704\ \text{g} + 58.44\ \text{g} = 158.144\ \text{g}$.

Step3: Determine significant figures

Volume of water: $100.\ \text{mL}$ (3 significant figures, the decimal shows precision).
Density: $0.99704\ \text{g/cm}^3$ (5 sig figs), NaCl mass: $58.44\ \text{g}$ (4 sig figs).
When adding, the result should have precision to the least precise decimal place.
$99.704$ (thousandths place) + $58.44$ (hundredths place) = $158.144$. The least precise is hundredths? Wait, no: $99.704$ has three decimal places, $58.44$ has two. So the sum should be rounded to the hundredths place? Wait, no—wait, $100.\ \text{mL}$: the decimal means the trailing zero is significant, so $100.$ has 3 sig figs. Density is 5, NaCl is 4. When multiplying for water mass: $0.99704 \times 100.$: $100.$ has 3 sig figs, so water mass is $99.7\ \text{g}$? Wait, no, wait: $100.\ \text{mL}$ is 3 sig figs (the decimal indicates that the zeros are significant). So $V = 100.\ \text{mL} = 1.00 \times 10^2\ \text{mL}$, 3 sig figs. Density is $0.99704\ \text{g/cm}^3$ (5 sig figs). So $m_{\text{water}} = 0.99704 \times 100. = 99.704\ \text{g}$, but since $100.$ has 3 sig figs, does that limit it? Wait, no—when multiplying, the number of sig figs is determined by the least number. But $100.$ is 3 sig figs, $0.99704$ is 5, so $m_{\text{water}}$ should be 3 sig figs? Wait, no, $100.\ \text{mL}$: the decimal is there to show that the two zeros are significant, so it's 3 sig figs. But density is 5, so when multiplying, $0.99704 \times 100. = 99.704$, but with 3 sig figs, that would be $99.7\ \text{g}$? Wait, no, maybe I messed up. Wait, the volume is $100.\ \text{mL}$—the decimal after the zero means that the measurement is precise to the milliliter, so it's 3 sig figs. But density is given as $0.99704\ \text{g/cm}^3$ (5 sig figs). So mass of water is $0.99704 \times 100. = 99.704\ \text{g}$. Now, when adding to $58.44\ \text{g}$ (4 sig figs, two decimal places), the sum's decimal places are determined by the least precise measurement. $99.704$ has three decimal places, $58.44$ has two. So we round to two decimal places? Wait, no: $99.704 + 58.44 = 158.144$. Now, check the significant figures from the original data:
- Volume: $100.\ \text{mL}$ (3 sig figs)
- Density: $0.99704\ \text{g/cm}^3$ (5 sig figs)
- NaCl mass: $58.44\ \text{g}$ (4 sig figs)

When calculating mass of water, the limiting factor is volume (3 sig figs), so $m_{\text{water}} = 99.7\ \text{g}$ (3 sig figs)? Wait, no, $0.99704 \times 100. = 99.704$, and $100.$ is 3 sig figs, so $99.704$ should be rounded to 3 sig figs: $99.7\ \text{g}$. Then adding $58.44\ \text{g}$: $99.7 + 58.44 = 158.14$, which rounds to $158.1\ \text{g}$? Wait, no, maybe I made a mistake. Wait, let's re-express:

Wait, the volume is $100.\ \text{mL}$—the decimal is important. In significant figures, $100.$ means three significant figures (the two zeros are significant because of the decimal). So $V = 1.00 \times 10^2\ \text{mL}$ (3 sig figs). Density is $0.99704\ \text{g/cm}^3$ (5 sig figs). So mass of water is $0.99704 \times 1.00 \times 10^2 = 99.704\ \text{g}$, but with 3 sig figs, that's $99.7\ \text{g}$? Wait, no, $1.00 \times 10^2$ has three sig figs, so the product should have three sig figs. So $0.99704 \times 100. = 99.704 \approx 99.7\ \text{g}$ (3 sig figs). Then NaCl is $58.44\ \text{g}$ (4 sig figs). Now, adding $99.7 + 58.44 = 158.14$. Now, the least number of decimal places in the terms being added: $99.7$ has one decimal place, $58.44$ has two. So we round to one decimal place? Wait, no: $99.7$ is to the tenths place, $58.44$ to the hundredths. When adding, the result should be rounded to the least precise decimal place, which is tenths (from $99.7$). So $158.14$ rounded to tenths is $158.1\ \text{g}$. Wait, but let's check the original numbers again. Wait, maybe the volume is $100.\ \text{mL}$ (three sig figs), density is $0.99704$ (five), so mass of water is $0.99704 \times 100. = 99.704\ \text{g}$ (since $100.$ has three sig figs, but the density has more, so maybe the volume's sig figs are three, but when multiplying, $0.99704 \times 100. = 99.704$, and since $100.$ is three sig figs, we can keep it as $99.704$ (maybe the decimal in $100.$ is just to show precision, not limiting sig figs? Wait, no—$100.$ has three sig figs, $100$ (without decimal) has one. So $100.$ is three. So $0.99704 \times 100. = 99.704$, which is three sig figs? No, $0.99704$ is five, $100.$ is three, so the product should have three sig figs: $99.7\ \text{g}$. Then adding $58.44\ \text{g}$: $99.7 + 58.44 = 158.14$. Now, $99.7$ has one decimal place, $58.44$ has two. So the sum should be rounded to one decimal place: $158.1\ \text{g}$. Wait, but let's check the answer options. The options include $158.1\ \text{g}$. Alternatively, maybe the volume is considered as $100.\ \text{mL}$ (three sig figs), but density is $0.99704$ (five), so mass of water is $99.704\ \text{g}$ (keeping more digits for calculation, then rounding at the end). So total mass is $99.704 + 58.44 = 158.144\ \text{g}$. Now, check the significant figures from the original data:

- Volume: $100.\ \text{mL}$ (3 sig figs)
- Density: $0.99704\ \text{g/cm}^3$ (5 sig figs)
- NaCl mass: $58.44\ \text{g}$ (4 sig figs)

When adding, the number of decimal places is determined by the least precise measurement. $99.704$ has three decimal places, $58.44$ has two. So we round to two decimal places? But $158.144$ rounded to two decimal places is $158.14\ \text{g}$. Wait, but $100.\ \text{mL}$: the decimal is there, so maybe it's three sig figs, but when calculating mass, $0.99704 \times 100. = 99.704$, which is three sig figs? No, $100.$ is three sig figs, so the mass of water is $99.7\ \text{g}$ (three sig figs). Then adding $58.44\ \text{g}$ (four sig figs), the sum is $158.14\ \text{g}$, but with three sig figs? Wait, no, significant figures for addition: it's about decimal places, not total sig figs. So $99.7$ (one decimal place) + $58.44$ (two decimal places) = $158.14$, which should be rounded to one decimal place: $158.1\ \text{g}$. Let's check the answer options. The options are:

- 158.144 g
- 158.14 g
- 158.1 g
- 158 g
- 1.6 × 10² g

Now, let's recalculate without rounding intermediate steps. Mass of water: $0.99704\ \text{g/cm}^3 \times 100.\ \text{mL} = 99.704\ \text{g}$ (since $1\ \text{mL} = 1\ \text{cm}^3$). Mass of NaCl: $58.44\ \text{g}$. Total mass: $99.704 + 58.44 = 158.144\ \text{g}$. Now, check the significant figures:

- Volume: $100.\ \text{mL}$ (3 sig figs)
- Density: $0.99704\ \text{g/cm}^3$ (5 sig figs)
- NaCl: $58.44\ \text{g}$ (4 sig figs)

The volume has 3 sig figs, so the mass of water is limited by that? Wait, no—$100.\ \text{mL}$ is 3 sig figs, but density is 5, so when multiplying, the result should have 3 sig figs: $99.7\ \text{g}$. Then adding $58.44\ \text{g}$: $99.7 + 58.44 = 158.14$, which is 158.1 when rounded to 4 sig figs? Wait, maybe the key is that $100.\ \text{mL}$ has three significant figures, but the decimal indicates that it's precise to the milliliter, so the volume is $100.0\ \text{mL}$? No, the problem says $100.\ \text{mL}$ (with a decimal after the zero). So $100.\ \text{mL}$ is three sig figs. Then, the mass of water is $0.99704 \times 100. = 99.704\ \text{g}$. Now, adding to $58.44\ \text{g}$, the total is $158.144\ \text{g}$. Now, we need to report to the proper number of significant figures. Let's check the least number of significant figures in the measurements:

- Volume: 3 sig figs
- Density: 5 sig figs
- NaCl: 4 sig figs

So the limiting factor is volume (3 sig figs)? No, when adding, significant figures are about decimal places, not total sig figs. Wait, no—when adding, the rule is that the result has the same number of decimal places as the term with the least number of decimal places.

- Mass of water: $99.704\ \text{g}$ (three decimal places)
- Mass of NaCl: $58.44\ \text{g}$ (two decimal places)

So the least number of decimal places is two (from NaCl). Therefore, we round $158.144\ \text{g}$ to two decimal places: $158.14\ \text{g}$. Wait, but $99.704$ has three decimal places, $58.44$ has two. So when adding, we go by the least decimal places, which is two. So $158.144$ rounded to two decimal places is $158.14\ \text{g}$. But let's check the options. $158.14\ \text{g}$ is an option. Alternatively, maybe the volume is $100.\ \text{mL}$ (three sig figs), so the total mass should have three sig figs? Wait, $158.144$ with three sig figs is $158\ \text{g}$? No, $158.144$ rounded to three sig figs is $158\ \text{g}$? Wait, no: $158.144$—the first three sig figs are 1, 5, 8, the next digit is 1, which is less than 5, so it stays $158\ \text{g}$. But that contradicts the decimal places. Wait, I think I confused the rules. Let's recall:

- For multiplication/division: result has same number of sig figs as least precise measurement.
- For addition/subtraction: result has same number of decimal places as least precise measurement.

So here, we have two steps:

1. Multiply density (5 sig figs) by volume (3 sig figs): $m_{\text{water}} = 0.99704 \times 100. = 99.704\ \text{g}$. Since volume has 3 sig figs, $m_{\text{water}}$ should have 3 sig figs: $99.7\ \text{g}$ (rounded to three sig figs).

2. Add $m_{\text{water}}$ (99.7 g, one decimal place) to $m_{\text{NaCl}}$ (58.44 g, two decimal places). The least number of decimal places is one, so the sum should have one decimal place: $99.7 + 58.44 = 158.14 \approx 158.1\ \text{g}$.

Ah, that makes sense. So step 1: multiply (3 sig figs) → $99.7\ \text{g}$. Step 2: add (one decimal place) → $158.1\ \text{g}$. So the correct answer is $158.1\ \text{g}$.

Answer:

158.1 g