Solutions containing 23 g HCOOH is/are :

Options

A. `46 g " of " 70%((w)/(V))HCOOH(d_("solution" )=1.40g//mL)`
B. `50 g " of 10 M " HCOOH(d_("solution" )=1g//mL)`
C. `50 g " of " 25 % ((w)/(w)) HCOOH`
D. 46 g " of 5 M " HCOOH`(d_(solution)=1 g//mL)`

Answer

To solve the question regarding solutions containing 23 g of formic acid (HCOOH), we will analyze each option provided in the question step by step.

### Step 1: Analyze the first option (46 g of 70% weight by volume HCOOH)
1. **Understanding 70% weight by volume**: This means that in 100 mL of solution, there are 70 g of HCOOH.
2. **Calculate the mass of the solution**: Given that the density of the solution is 1.4 g/mL, we can find the mass of 100 mL of solution:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 1.4 \, \text{g/mL} \times 100 \, \text{mL} = 140 \, \text{g} \]
3. **Calculate the amount of HCOOH in 46 g of this solution**:
\[ \text{Amount of HCOOH in 1 g of solution} = \frac{70 \, \text{g}}{140 \, \text{g}} = 0.5 \, \text{g} \]
Therefore, in 46 g of the solution:
\[ \text{Amount of HCOOH} = 0.5 \, \text{g/g} \times 46 \, \text{g} = 23 \, \text{g} \]
This means this option contains 23 g of HCOOH.

### Step 2: Analyze the second option (50 g of 10 molar HCOOH)
1. **Understanding molarity**: Molarity (M) is defined as moles of solute per liter of solution. A 10 M solution means there are 10 moles of HCOOH in 1 L of solution.
2. **Calculate the number of moles in 50 g of solution**: The molecular weight of HCOOH is 46 g/mol. The number of moles in 50 g is:
\[ \text{Number of moles} = \frac{50 \, \text{g}}{46 \, \text{g/mol}} \approx 1.087 \, \text{moles} \]
3. **Calculate the volume of the solution**: Since the solution is 10 M:
\[ \text{Volume} = \frac{\text{Number of moles}}{\text{Molarity}} = \frac{1.087 \, \text{moles}}{10 \, \text{mol/L}} = 0.1087 \, \text{L} = 108.7 \, \text{mL} \]
4. **Calculate the mass of HCOOH in this volume**:
\[ \text{Mass of HCOOH} = \text{Number of moles} \times \text{Molecular weight} = 1.087 \, \text{moles} \times 46 \, \text{g/mol} \approx 50 \, \text{g} \]
Since the solution is 10 M, we find:
\[ \text{Mass of HCOOH} = 10 \, \text{mol/L} \times 0.1087 \, \text{L} \times 46 \, \text{g/mol} \approx 23 \, \text{g} \]
This means this option also contains 23 g of HCOOH.

### Step 3: Analyze the third option (50 g of 25% weight by weight HCOOH)
1. **Understanding 25% weight by weight**: This means that in 100 g of solution, there are 25 g of HCOOH.
2. **Calculate the amount of HCOOH in 50 g of this solution**:
\[ \text{Amount of HCOOH} = \frac{25 \, \text{g}}{100 \, \text{g}} \times 50 \, \text{g} = 12.5 \, \text{g} \]
This is not equal to 23 g, so this option is incorrect.

### Step 4: Analyze the fourth option (46 g of 5 molar HCOOH)
1. **Understanding molarity**: A 5 M solution means there are 5 moles of HCOOH in 1 L of solution.
2. **Calculate the number of moles in 46 g of solution**:
\[ \text{Number of moles} = \frac{46 \, \text{g}}{46 \, \text{g/mol}} = 1 \, \text{mole} \]
3. **Calculate the volume of the solution**:
\[ \text{Volume} = \frac{1 \, \text{mole}}{5 \, \text{mol/L}} = 0.2 \, \text{L} = 200 \, \text{mL} \]
4. **Calculate the mass of HCOOH in this volume**:
\[ \text{Mass of HCOOH} = 5 \, \text{mol/L} \times 0.2 \, \text{L} \times 46 \, \text{g/mol} = 46 \, \text{g} \]
This is not equal to 23 g, so this option is also incorrect.

### Conclusion
The correct options are the first and second options, which both contain 23 g of HCOOH.