3. If 5.20 g of HCl is added to enough distilled water to form 3.00 L of solution, what is the molarity of the solution? [Use molecular weight: HCl, 36.46 amu]
5.20/36.46= 0.1426 X 3.00=0.428 M
IS THIS CORRECT
part right; part wrong; overall is wrong.
Yes, mols = grams/molar mass = 0.1426
Then M = mols/L = 0.1426/3.00 L = ?
Then round to 3 significant figures.
To determine the molarity of the HCl solution, we need to calculate the moles of HCl and then divide it by the volume of the solution.
First, let's calculate the number of moles of HCl:
Moles = Mass / Molecular weight
Given that the mass of HCl is 5.20 grams and the molecular weight of HCl is 36.46 amu, we can plug these values into the formula:
Moles = 5.20 g / 36.46 g/mol ≈ 0.1426 mol
Next, we need to calculate the molarity of the solution:
Molarity = Moles / Volume
Given that the volume of the solution is 3.00 L, we can plug in the values:
Molarity = 0.1426 mol / 3.00 L = 0.0475 M
Therefore, the molarity of the solution is approximately 0.0475 M.
Looking at the calculation steps you provided, it seems that you correctly calculated the moles of HCl. However, you multiplied it by 3, instead of dividing it by 3, to find the molarity. So your final answer of 0.428 M is incorrect.
The correct molarity is 0.0475 M.