Calculate the volume in milliliters of a M aluminum chloride solution that contains of aluminum chloride . Round your answer to significant digits.

Respuesta :

Answer:

326.9mL

Explanation:

...Of a 1.3M aluminium chloride solution that contains 425mmol of aluminium chloride...

The prefix milli, m, means denotes factor of one thousandth (1x10⁻³), that means 425mmol are:

425mmol * (1x10⁻³mol / 1mmol) = 0.425moles of AlCl₃.

In the same way, Molarity, m, is defined as moles of solute (aluminium chloride) per liter. The 1.3M contains 1.3 moles per liter. There are 0.425moles in:

0.425moles * (1L / 1.3moles) = 0.3269L.

In milliliters:

0.3269L * (1000mL / 1L) =

326.9mL