<u>Given:</u>
Mass of MgBr2 = 0.500 g
<u>To determine:</u>
Number of anions in 0.500 g MgBr2
<u>Explanation:</u>
Molar mass of MgBr2 = 24 + 2 (80) = 184 g/mol
Moles of MgBr2 = 0.500 g/184 g.mol-1 = 0.00271 moles
Based on stoichiometry-
1 mole of MgBr2 has 1 mole of Mg2+ cations and 2 moles of Br- anions
Therefore, 0.00271 moles of MgBr2 will have: 2 * 0.00271 = 0.00542 moles of Br-
Now,
1 mole of Br- contains 6.023 * 10²³ anions
0.00542 moles of Br- contain: 0.00542 * 6.023*10²³ = 3.264*10²¹ anions
Ans: There are 3.264*10²¹ anions in 0.5 g of MgBr2
Molarity can be defined as the number of moles of substance dissolved in 1 L of solution.
In the given question ,
number of LiOH moles - 1.495 mol
Dissolved volume - 750 mL
molarity is calculated for 1 L = 1000 mL
In 750 mL - 1.495 mol of LiOH is dissolved
Therefore in 1000 mL - 1.495 mol / 750 mL x 1000 = 1.99 mol
