Chemistry, asked by gauraverahkp9fmz3, 11 months ago

Calculate the number of moles of HCl present in 500 mL of 0.1M HCl solution.​

Answers

Answered by ExpertSohanXLR8
36

Explanation:

volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 500 mL into liters by using the conversion factor 1000mL = 1L. When 500 mL is divided by 1000 mL/L, we obtain a volume of 0.50 L.

Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:

Moles of solute = (liters of solution) * (Molarity)

Moles of solute = (0.50 L) (6.0 M) = 3.0 mol of HCl

Now we have to convert the 3.0 mol of HCl into grams of HCl. This can be done by multiplying 3.0 mol by the molecular weight of HCl, which is 36.46 g/mol.

(3.0 mol)(36.46 g/mol) = 109 g HCl

Answered by visruthaa2609
17

Answer:

Explanation:

volume of solution. However, the volume does not have the proper units since it is given in terms of milliliters instead of liters. We can convert 500 mL into liters by using the conversion factor 1000mL = 1L. When 500 mL is divided by 1000 mL/L, we obtain a volume of 0.50 L.

Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:

Moles of solute = (liters of solution) * (Molarity)

Moles of solute = (0.50 L) (6.0 M) = 3.0 mol of HCl

Now we have to convert the 3.0 mol of HCl into grams of HCl. This can be done by multiplying 3.0 mol by the molecular weight of HCl, which is 36.46 g/mol.

(3.0 mol)(36.46 g/mol) = 109 g HCl

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