Question

A chemist adds 1.80L of a 1.1/molL aluminum chloride AlCl3 solution to a reaction flask. Calculate the millimoles of aluminum chloride the chemist has added to the flask. Round your answer to 2 significant digits.

Answer 1

Answer:

2000 millimoles of AlCl₃

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 1.8 L

Molarity of solution = 1.1 mol /L

Millmole of AlCl₃ =?

Next, we shall determine the number of mole of AlCl₃ in the solution.

This can be obtained as follow:

Volume of solution = 1.8 L

Molarity of AlCl₃ solution = 1.1 mol /L

Number of mole of AlCl₃ =?

Molarity = mole /Volume

1.1 = Number of mole of AlCl₃ / 1.8

Cross multiply

Number of mole of AlCl₃ = 1.1 × 1.8

Number of mole of AlCl₃ = 1.98 moles

Finally, we shall convert 1.98 moles to millimoles. This can be obtained as follow:

1 mole = 1000 millimoles

Therefore,

1.98 mole = 1.98 mole × 1000 millimoles / 1 mole

1.98 mole = 1980 millimoles

1.98 mole ≈ 2000 millimoles

Thus, the chemist added 2000 millimoles of AlCl₃