Calculate the volume in liters of a M mercury(II) iodide solution that contains of mercury(II) iodide . Round your answer to significant digits.

41L

Explanation:

Of a 4.8x10⁻⁵M mercury (II) iodide that contains 900mg of mercury (II) iodide. 2 significant digits

Molarity, M, is an unit of concentration in chemistry defined as the ratio between moles of solute (Mercury (II) iodide in this case) per Liter of solution.

A 4.8x10⁻⁵M solution contains 4.8x10⁻⁵ moles of solute per liter.

Now, 900mg = 0.900g of mercury (II) iodide (Molar mass: 454.4g/mol) are:

0.900g × (1mol / 454.4g) = 1.98x10⁻³moles of HgI₂

If in 1L there are 4.8x10⁻⁵ moles of HgI₂, There are 1.98x10⁻³moles of HgI₂ in:

1.98x10⁻³moles of HgI₂ ₓ (1L / 4.8x10⁻⁵moles) =

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )