Common salt obtained from sea-water contains `8.775% Na Cl` by mass. The number of formula units of `NaCl` present in `25 g` of this salt is :

Options

A. `3.367 xx 10 ^(23)"formula units"`
B. `2.258 xx 10 ^(22)"formula units"`
C. `3.176 xx 10 ^(23)"formula units"`
D. `4.73 xx 10 ^(25)"formula units"`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the mass of NaCl in 25 g of salt
Given that the common salt contains 8.775% NaCl by mass, we can calculate the mass of NaCl in 25 g of salt.

\[ \text{Mass of NaCl} = \left( \frac{8.775}{100} \right) \times 25 \, \text{g} \]

\[ \text{Mass of NaCl} = 0.08775 \times 25 = 2.19375 \, \text{g} \]

### Step 2: Calculate the number of moles of NaCl
Next, we need to find the number of moles of NaCl using its molar mass. The molar mass of NaCl is calculated as follows:

\[ \text{Molar mass of NaCl} = \text{Atomic mass of Na} + \text{Atomic mass of Cl} = 23 \, \text{g/mol} + 35.5 \, \text{g/mol} = 58.5 \, \text{g/mol} \]

Now, we can calculate the number of moles of NaCl:

\[ \text{Number of moles of NaCl} = \frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}} = \frac{2.19375 \, \text{g}}{58.5 \, \text{g/mol}} \]

\[ \text{Number of moles of NaCl} \approx 0.0375 \, \text{mol} \]

### Step 3: Calculate the number of formula units of NaCl
Using Avogadro's number, which is \(6.022 \times 10^{23}\) formula units/mol, we can find the number of formula units of NaCl:

\[ \text{Number of formula units of NaCl} = \text{Number of moles of NaCl} \times 6.022 \times 10^{23} \]

\[ \text{Number of formula units of NaCl} = 0.0375 \, \text{mol} \times 6.022 \times 10^{23} \, \text{formula units/mol} \]

\[ \text{Number of formula units of NaCl} \approx 2.258 \times 10^{22} \, \text{formula units} \]

### Final Answer:
The number of formula units of NaCl present in 25 g of common salt is approximately \(2.258 \times 10^{22}\) formula units.

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