Question

a chemistry student weighs out 0.160 kg of an unknown solid compound x and adds it to 700. ml of distilled water at 22. °c. after 10 minutes of stirring, only some of the x has dissolved. the student drains off the solution, then washes, dries and weighs the x that did not dissolve. it weighs 0.048 kg.
using only the information above, can you calculate the solubility of x in water at 22. °c?
○ yes
○ no
if you said yes, calculate it.
be sure your answer has a unit symbol and the right number of significant digits.

Explanation:

Step1: Determine dissolved mass

The initial mass of \( X \) is \( 0.160 \, \text{kg} \), and the undissolved mass is \( 0.048 \, \text{kg} \). So, the mass of \( X \) that dissolved is \( m_{\text{dissolved}} = 0.160 - 0.048 = 0.112 \, \text{kg} \).

Step2: Volume of water

The volume of water is \( 700. \, \text{mL} \). Since the density of water is approximately \( 1 \, \text{kg/L} \) (or \( 1 \, \text{g/mL} \)), the mass of water is \( m_{\text{water}} = 700. \, \text{mL} \times 1 \, \text{g/mL} = 700 \, \text{g} = 0.700 \, \text{kg} \).

Step3: Calculate solubility

Solubility (\( S \)) is the mass of solute dissolved per mass (or volume) of solvent. Using mass of solvent (water), \( S = \frac{m_{\text{dissolved}}}{m_{\text{water}}} \times 1000 \, \text{g/kg} \) (to get g/kg, a common solubility unit). Substituting values: \( S = \frac{0.112 \, \text{kg}}{0.700 \, \text{kg}} \times 1000 \, \text{g/kg} = 160 \, \text{g/kg} \) (or \( 0.160 \, \text{kg/kg} \), but g/kg is more standard for solubility here). Wait, actually, let's check units again. If we use grams: \( 0.112 \, \text{kg} = 112 \, \text{g} \), \( 0.700 \, \text{kg} = 700 \, \text{g} \). Then \( S = \frac{112 \, \text{g}}{700 \, \text{g}} \times 100 \, \text{g/100g water} \)? Wait, no, solubility can be in g per 100 g water. Let's recalculate: \( \frac{112 \, \text{g}}{700 \, \text{g water}} = \frac{x}{100 \, \text{g water}} \), so \( x = \frac{112 \times 100}{700} = 16.0 \, \text{g/100 g water} \). Wait, earlier step had a unit mix - let's correct. The dissolved mass is \( 0.160 - 0.048 = 0.112 \, \text{kg} = 112 \, \text{g} \). Water is \( 700 \, \text{mL} = 700 \, \text{g} \) (since density of water is 1 g/mL). So solubility in g per 100 g water: \( \frac{112 \, \text{g}}{700 \, \text{g}} \times 100 = 16.0 \, \text{g/100 g water} \). Or in g/L: \( \frac{112 \, \text{g}}{0.700 \, \text{L}} = 160 \, \text{g/L} \). But let's use the correct approach. The key is: dissolved mass = 0.112 kg = 112 g, water volume = 700 mL = 0.700 L (since 1 L = 1000 mL). So solubility (g/L) is \( \frac{112 \, \text{g}}{0.700 \, \text{L}} = 160 \, \text{g/L} \). Wait, but let's check significant figures. Initial masses: 0.160 (3 sig figs), 0.048 (2 sig figs), volume 700. (3 sig figs). The dissolved mass: 0.160 - 0.048 = 0.112 (the subtraction: 0.160 has three decimal places? No, 0.160 is 160 mg? Wait, no, 0.160 kg is 160 grams, 0.048 kg is 48 grams. So 160 - 48 = 112 grams (exact subtraction here, since 0.160 has three sig figs, 0.048 has two, but subtraction is about decimal places. 0.160 kg is 160. g (three sig figs), 0.048 kg is 48. g (two sig figs). So 160. - 48. = 112. g (the decimal places: 160. has one decimal place? No, 0.160 kg is 160 grams (three sig figs), 0.048 kg is 48 grams (two sig figs). So the result of subtraction is 112 grams (the least number of decimal places? Wait, no, subtraction is about the number of decimal places in the original numbers. 0.160 kg is 160. g (three sig figs, decimal at the end for sig fig), 0.048 kg is 48 g (two sig figs). So 160. - 48 = 112 g (the uncertainty is in the ones place? Maybe. Then volume is 700. mL (three sig figs), so 0.700 kg (three sig figs). Then solubility: 112 g / 0.700 kg = 160 g/kg (three sig figs? Wait, 112 has three, 0.700 has three, so the result should have three. 112 / 0.700 = 160. (exactly 160, but with three sig figs, it's 1.60 × 10² g/kg, or 160. g/kg). Wait, maybe the correct unit is g per 100 g water. Let's do that: 112 g solute / 700 g water = (112 / 700) × 100 g per 100 g water = 16.0 g per 100 g water. Yes, that makes sense. So solubility is 16.0 g/100 g water (or 160 g/kg, or 160 g/L since water is 1 g/mL).

Answer:

The solubility of \( X \) in water at \( 22^\circ \text{C} \) is \( \boldsymbol{16.0 \, \text{g/100 g water}} \) (or \( \boldsymbol{160 \, \text{g/kg}} \), or \( \boldsymbol{160 \, \text{g/L}} \), depending on unit convention, with appropriate significant figures).