A student dissolves 45.2g of mg(OH)2(S) in enough water to make 1.00 L of solution. The concentration of the solution is
To find the concentration of the solution, we first need to calculate the molar mass of magnesium hydroxide (Mg(OH)2).
Molar mass of Mg(OH)2 = 24.3 (Mg) + 2(16.0) (O) + 2(1.0) (H)
= 24.3 + 32 + 2
= 58.3 g/mol
Next, we calculate the number of moles of Mg(OH)2 in 45.2g:
Number of moles = Mass/Molar mass
Number of moles = 45.2g / 58.3 g/mol
Number of moles = 0.775 mol
Since the solution is 1.00 L in volume, the concentration of Mg(OH)2 is:
Concentration = Number of moles / Volume
Concentration = 0.775 mol / 1.00 L
Concentration = 0.775 mol/L
Therefore, the concentration of the solution is 0.775 mol/L.