The density of liquid (mol.wt. `= 70`) is `1.2 g mL^(-1)`. If `2 mL` of liquid contains `35` drops, the number of molecules of liquid in one drop are:

Options

A. `((1.2)/(35))N_(A)`
B. `(1)/(1.2)((1)/(35))^(2)N_(A)`
C. `(1.2)/((35)^(2))N_(A)`
D. `1.2N_(A)`

Explanation

To solve the problem, we need to find the number of molecules of the liquid in one drop. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the weight of 2 mL of the liquid
The density of the liquid is given as \(1.2 \, \text{g/mL}\). Therefore, the weight of 2 mL of the liquid can be calculated as follows:

\[ \text{Weight of 2 mL} = \text{Density} \times \text{Volume} = 1.2 \, \text{g/mL} \times 2 \, \text{mL} = 2.4 \, \text{g} \]

### Step 2: Calculate the weight of one drop
We know that 2 mL of the liquid contains 35 drops. To find the weight of one drop, we divide the total weight of the liquid by the number of drops:

\[ \text{Weight of one drop} = \frac{\text{Weight of 2 mL}}{\text{Number of drops}} = \frac{2.4 \, \text{g}}{35} \approx 0.06857 \, \text{g} \]

### Step 3: Calculate the number of moles in one drop
Next, we need to convert the weight of one drop into moles. We use the molecular weight of the liquid, which is given as 70 g/mol. The number of moles in one drop can be calculated using the formula:

\[ \text{Number of moles in one drop} = \frac{\text{Weight of one drop}}{\text{Molecular weight}} = \frac{0.06857 \, \text{g}}{70 \, \text{g/mol}} \approx 0.000978 \, \text{mol} \]

### Step 4: Calculate the number of molecules in one drop
To find the number of molecules in one drop, we use Avogadro's number (\(N_A \approx 6.022 \times 10^{23} \, \text{molecules/mol}\)):

\[ \text{Number of molecules in one drop} = \text{Number of moles} \times N_A = 0.000978 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 5.88 \times 10^{20} \, \text{molecules} \]

### Final Answer
The number of molecules of liquid in one drop is approximately \(5.88 \times 10^{20}\).

---