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The mole fraction of NaCl in a solution ...

The mole fraction of NaCl in a solution containing 1 mole of NaCl in 1000 g of water is

A

`0.0177`

B

`0.001`

C

`0.5`

D

`0.244`

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The correct Answer is:
To find the mole fraction of NaCl in a solution containing 1 mole of NaCl in 1000 g of water, we will follow these steps: ### Step 1: Identify the number of moles of NaCl We are given that there is 1 mole of NaCl in the solution. - **Moles of NaCl (n_NaCl)** = 1 mole ### Step 2: Calculate the number of moles of water (H2O) To find the moles of water, we need to use the formula: \[ \text{Moles of H2O} = \frac{\text{mass of H2O (in grams)}}{\text{molar mass of H2O (in g/mol)}} \] The molar mass of water (H2O) is calculated as follows: - Hydrogen (H) has a molar mass of approximately 1 g/mol, and there are 2 hydrogen atoms. - Oxygen (O) has a molar mass of approximately 16 g/mol. \[ \text{Molar mass of H2O} = (2 \times 1) + 16 = 18 \text{ g/mol} \] Now, we can calculate the moles of water: \[ \text{Moles of H2O} = \frac{1000 \text{ g}}{18 \text{ g/mol}} \approx 55.56 \text{ moles} \] ### Step 3: Calculate the total number of moles in the solution Now we can find the total number of moles in the solution: \[ \text{Total moles} = n_{NaCl} + n_{H2O} = 1 + 55.56 \approx 56.56 \text{ moles} \] ### Step 4: Calculate the mole fraction of NaCl The mole fraction of NaCl (x_NaCl) is given by the formula: \[ x_{NaCl} = \frac{n_{NaCl}}{n_{NaCl} + n_{H2O}} \] Substituting the values we have: \[ x_{NaCl} = \frac{1}{1 + 55.56} = \frac{1}{56.56} \approx 0.0177 \] ### Final Answer: The mole fraction of NaCl in the solution is approximately **0.0177**. ---

To find the mole fraction of NaCl in a solution containing 1 mole of NaCl in 1000 g of water, we will follow these steps: ### Step 1: Identify the number of moles of NaCl We are given that there is 1 mole of NaCl in the solution. - **Moles of NaCl (n_NaCl)** = 1 mole ### Step 2: Calculate the number of moles of water (H2O) To find the moles of water, we need to use the formula: ...
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