How much BaCl2 would be needed to make 250 mL of a solution having same concentration of Cl^- as the one containing 3.78g of NaCl per 100 mL?
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To find out how much BaCl2 is needed to make a solution with the same concentration of Cl^- as the NaCl solution, you need to use stoichiometry and mole ratios.
First, calculate the molar mass of NaCl and BaCl2:
- Molar mass of NaCl (Na = 22.990 g/mol, Cl = 35.453 g/mol):
Molar mass of NaCl = 22.990 g/mol + 35.453 g/mol = 58.443 g/mol
- Molar mass of BaCl2 (Ba = 137.33 g/mol, Cl = 35.453 g/mol):
Molar mass of BaCl2 = 137.33 g/mol + (2 * 35.453 g/mol) = 208.253 g/mol
Next, calculate the number of moles of NaCl in the given solution:
- Given: 3.78 g of NaCl per 100 mL
- Convert mL to L: 100 mL = 0.1 L
- Calculate the number of moles using the formula: moles = mass / molar mass
- Moles of NaCl = (3.78 g / 58.443 g/mol) * 0.1 L = 0.06472 mol
Now, we can determine the number of moles of Cl^- ions present in the NaCl solution. Since NaCl dissociates into one Na+ ion and one Cl^- ion when it dissolves, the number of moles of Cl^- ions is the same as the number of moles of NaCl.
Therefore, the number of moles of Cl^- ions = 0.06472 mol.
Now, let's calculate the volume in liters for the desired BaCl2 solution:
- Given volume = 250 mL = 0.25 L
Finally, we can calculate the mass of BaCl2 needed:
- Since BaCl2 dissociates into one Ba^2+ ion and two Cl^- ions when it dissolves, the number of moles of Cl^- ions would be half the number of moles of BaCl2.
- Therefore, moles of BaCl2 required = moles of Cl^- ions / 2 = 0.06472 mol / 2 = 0.03236 mol.
The mass of BaCl2 required can be calculated using the formula: mass = moles x molar mass
- Mass of BaCl2 required = 0.03236 mol x 208.253 g/mol = 6.747 g
So, approximately 6.747 grams of BaCl2 would be needed to make 250 mL of the BaCl2 solution with the same concentration of Cl^- as the NaCl solution containing 3.78 grams of NaCl per 100 mL.
16.8 gm
3.78 g NaCl/100 mL = 37.8 g/L solution and that's how many mols.
37.8/molar mass = approx 0.6 mols/L but you need answer than this estimate. This makes (Cl^-) = approx 0.6M
You want to make 250 mL of BaCl2 and since BaCl2 has 2 Cl atoms/1 Bacl2 molecule, you only need 0.3 M; therefore,
mols BaCl2 needed = M x L = ?
grams BaCl2 = mols BaCl2 x molar mass BaCl2.