A 10.0 g sample of bismuth tribromide,BiBr₃,contains: A)5.37 × 10²² total number of ions B)0.322 mol BiBr₃ C)3.14 × 10²² formula units BiBr₃ D)1.34 × 10²² bromide ions E)4.020 × 10²² total number of ions

Question

Quizplus · Accepted Answer

To find the total number of ions, we need to first determine the formula for BiBr3. Bismuth has a 3+ charge, while bromine has a 1- charge, so we need 3 bromide ions to balance the charge of 1 bismuth ion. Therefore, the formula for BiBr3 is BiBr3.

Next, we need to determine the number of moles of BiBr3 in the sample:
0.0100 g BiBr3 × (1 mol BiBr3/394.692 g BiBr3) = 2.53 × 10^-5 mol BiBr3

Now we can use Avogadro's number to find the total number of ions:
2.53 × 10^-5 mol BiBr3 × (6.02 × 10^23 ions/mol) = 5.37 × 10^18 ions

A 10.0 G Sample of Bismuth Tribromide,BiBr3,contains

A 10.0 G Sample of Bismuth Tribromide,BiBr₃,contains