### Step-by-Step Solution:
#### 1. Which has a higher molecular weight: [tex]\( \text{NaCl} \)[/tex] or [tex]\( \text{CaCl}_2 \)[/tex]?
To determine the molecular weights, we sum the atomic weights of the elements in each compound:
- For [tex]\( \text{NaCl} \)[/tex]:
- Sodium (Na) has an atomic weight of 23.
- Chlorine (Cl) has an atomic weight of 35.5.
- Hence, the molecular weight of NaCl is [tex]\( 23 + 35.5 = 58.5 \)[/tex] g/mol.
- For [tex]\( \text{CaCl}_2 \)[/tex]:
- Calcium (Ca) has an atomic weight of 40.
- Chlorine (Cl) has an atomic weight of 35.5, and there are two chlorine atoms.
- Hence, the molecular weight of [tex]\( \text{CaCl}_2 \)[/tex] is [tex]\( 40 + 2 \times 35.5 = 111 \)[/tex] g/mol.
Therefore, [tex]\( \text{CaCl}_2 \)[/tex] has a higher molecular weight of 111 g/mol compared to [tex]\( \text{NaCl} \)[/tex] which has a molecular weight of 58.5 g/mol.
#### 2. How many ions does [tex]\( \text{NaCl} \)[/tex] dissociate into? How many ions does [tex]\( \text{CaCl}_2 \)[/tex] dissociate into?
- [tex]\( \text{NaCl} \)[/tex] dissociates into:
- One sodium ion (Na[tex]\(^+\)[/tex])
- One chloride ion (Cl[tex]\(^-\)[/tex])
- Total: [tex]\( 2 \)[/tex] ions.
- [tex]\( \text{CaCl}_2 \)[/tex] dissociates into:
- One calcium ion (Ca[tex]\(^2+\)[/tex])
- Two chloride ions (2Cl[tex]\(^-\)[/tex])
- Total: [tex]\( 3 \)[/tex] ions.
#### 3. About how many moles of [tex]\( \text{NaCl} \)[/tex] are in a cc (approximately 1.1 g) of [tex]\( \text{NaCl} \)[/tex]?
To find the moles of [tex]\( \text{NaCl} \)[/tex], we use the formula:
[tex]\[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \][/tex]
For [tex]\( \text{NaCl} \)[/tex]:
- Mass = 1.1 g
- Molar Mass = 58.5 g/mol
Hence,
[tex]\[ \text{Moles of NaCl} = \frac{1.1 \text{ g}}{58.5 \text{ g/mol}} \approx 0.0188 \text{ moles} \][/tex]
#### 4. About how many moles of [tex]\( \text{CaCl}_2 \)[/tex] are in a cc (approximately 1.3 g) of [tex]\( \text{CaCl}_2 \)[/tex]?
Using the same formula, for [tex]\( \text{CaCl}_2 \)[/tex]:
- Mass = 1.3 g
- Molar Mass = 111 g/mol
Hence,
[tex]\[ \text{Moles of CaCl}_2 = \frac{1.3 \text{ g}}{111 \text{ g/mol}} \approx 0.0117 \text{ moles} \][/tex]
### Summary of Answers:
1. The compound with the higher molecular weight is [tex]\( \text{CaCl}_2 \)[/tex].
2.
- [tex]\( \text{NaCl} \)[/tex] dissociates into 2 ions.
- [tex]\( \text{CaCl}_2 \)[/tex] dissociates into 3 ions.
3. There are approximately 0.0188 moles of [tex]\( \text{NaCl} \)[/tex] in 1 cc (1.1 g) of [tex]\( \text{NaCl} \)[/tex].
4. There are approximately 0.0117 moles of [tex]\( \text{CaCl}_2 \)[/tex] in 1 cc (1.3 g) of [tex]\( \text{CaCl}_2 \)[/tex].
