Question

1. How many ATOMS of boron are present in 8.79 grams of boron trifluoride ?  atoms of boron .


2. How many GRAMS of fluorine are present in 5.10×10²²molecules of boron trifluoride ? grams of fluorine .

Answer 1

Boron Trifluoride : BF₃.
Molar mass of BF₃ : 67.82 g/mol

1. Moles of BF3 in 8.79 g : 8.79 g/ (67.82 g/mol​)
   : 0.1296 mole
Now 1 mole of BF₃ contains 6.022 x 10²³ molecules of BF₃.
Therefore, 0.1296 mole​ of BF₃ contains = (0.1296 mole​​) x (6.022 x 10²³ molecules​/ mole)
   = 7.8045 x 10²² molecules of BF₃.
Now 1 molecule of BF3 has 1 atom of Boron B, therefore 7.8045 x 10²² molecules of BF₃.​ will have 7.8045 x 10²² atoms of Boron.

2.) Determine the moles of BF3 = 5.10 x 10²² molecules/ (6.022 x 10²³ molecules/ mole)
   = 0.0847 mole BF3.
Now 1 mole of BF3 contains 3 moles Fluorine F. Therefore, 0.0847 mole BF3 will have = 3 x 0.0847 mole F
   = 0.2541 mole F
Now, molar mass of fluorine = 19 g/ mol
Grams of fluorine = (0.2541 mole) x (19 g/ mol)
   = 4.8279 g