Flocculation value for `As_2S_3` sol by HCl is 30 m mole `L^( –1)` . Calcualte mass of `H_2SO_4` required in gram for 250 mL sol.

Answer

To solve the problem of calculating the mass of \( H_2SO_4 \) required for a 250 mL solution given the flocculation value for \( As_2S_3 \) sol by HCl is 30 m mole \( L^{-1} \), we can follow these steps:

### Step 1: Understand the Flocculation Value
The flocculation value indicates the amount of electrolyte (in this case, HCl) required to coagulate a colloidal solution. Here, we know that 30 millimoles of HCl are needed per liter of the solution.

### Step 2: Determine the Equivalent Requirement for \( H_2SO_4 \)
Since \( H_2SO_4 \) dissociates into 2 \( H^+ \) ions, it can provide twice the amount of \( H^+ \) compared to HCl, which provides one \( H^+ \) ion. Therefore, to find the equivalent amount of \( H_2SO_4 \) needed, we can use the following relationship:

\[ \text{Millimoles of } H_2SO_4 = \frac{\text{Millimoles of HCl}}{2} = \frac{30 \text{ m moles}}{2} = 15 \text{ m moles} \]

### Step 3: Calculate the Amount for 250 mL
Since the flocculation value is given per liter, we need to adjust this for 250 mL:

\[ \text{Millimoles of } H_2SO_4 \text{ for 250 mL} = 15 \text{ m moles} \times \frac{250 \text{ mL}}{1000 \text{ mL}} = 3.75 \text{ m moles} \]

### Step 4: Convert Millimoles to Grams
Next, we need to convert millimoles of \( H_2SO_4 \) to grams. The molar mass of \( H_2SO_4 \) is 98 g/mol.

First, convert millimoles to moles:

\[ 3.75 \text{ m moles} = 3.75 \times 10^{-3} \text{ moles} \]

Now, calculate the mass:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} = 3.75 \times 10^{-3} \text{ moles} \times 98 \text{ g/mol} \]

\[ \text{Mass} = 0.3675 \text{ grams} \]

### Final Answer
The mass of \( H_2SO_4 \) required for 250 mL of the solution is **0.3675 grams**.

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