To find the mass of magnesium chlorate \((\text{Mg(ClO}_3\text{)}_2)\) in the solution, we can use the formula:
\[ \text{mass} = \text{molarity} \times \text{volume} \times \text{molar mass} \]
-
Convert volume from mL to L: \[ 300 \text{ mL} = 0.300 \text{ L} \]
-
Calculate the number of moles of magnesium chlorate: \[ \text{moles} = \text{molarity} \times \text{volume} = 4.02 , \text{mol/L} \times 0.300 , \text{L} = 1.206 , \text{mol} \]
-
Calculate the molar mass of magnesium chlorate \((\text{Mg(ClO}_3\text{)}_2)\):
- Atomic mass of Mg = 24.31 g/mol
- Atomic mass of Cl = 35.45 g/mol (there are 2 Cl atoms)
- Atomic mass of O = 16.00 g/mol (there are 6 O atoms, since it’s \((\text{ClO}_3)\) twice)
Now calculating the molar mass: \[ \text{Molar mass} = 24.31 + 2(35.45) + 6(16.00) \]
\[ = 24.31 + 70.90 + 96.00 \] \[ = 191.21 , \text{g/mol} \]
-
Calculate the mass of magnesium chlorate: \[ \text{mass} = \text{moles} \times \text{molar mass} = 1.206 , \text{mol} \times 191.21 , \text{g/mol} \] \[ \text{mass} = 230.57 , \text{g} \]
So, the mass of magnesium chlorate in the 300 ml sample is approximately 230.57 g.
Given the options (0.021 gr, 0.07 gr, 191.21 gr, and \(7 \times 10^{-5}\) gr), none of the provided answers match the calculated mass. Each option seems to lack correctness based on the assumed number of moles from the calculation.
The correct answer should round to 230.57 g, and no options correspond. Thus, there seems to be an inconsistency in the provided options.