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Answer :
To solve the problem of finding out how many grams of [tex]\( \text{MgCl}_2 \)[/tex] are needed to make 1.55 L of a 0.0520 M chloride ion solution, we can follow these steps:
1. Understand the relationship in the chemical formula: Magnesium chloride ([tex]\( \text{MgCl}_2 \)[/tex]) dissociates in water to form one magnesium ion ([tex]\( \text{Mg}^{2+} \)[/tex]) and two chloride ions ([tex]\( \text{Cl}^- \)[/tex]). This means that each molecule of [tex]\( \text{MgCl}_2 \)[/tex] contributes two chloride ions to the solution.
2. Determine the molarity of [tex]\( \text{MgCl}_2 \)[/tex]: Since the solution requires a 0.0520 M concentration of chloride ions, and each mole of [tex]\( \text{MgCl}_2 \)[/tex] produces two moles of chloride ions, the concentration of [tex]\( \text{MgCl}_2 \)[/tex] needs to be half of the chloride ion concentration. Therefore, the molarity of [tex]\( \text{MgCl}_2 \)[/tex] is:
[tex]\[
\text{Molarity of } \text{MgCl}_2 = \frac{0.0520 \, \text{M}}{2} = 0.0260 \, \text{M}
\][/tex]
3. Calculate the moles of [tex]\( \text{MgCl}_2 \)[/tex] needed: Use the molarity formula, which is:
[tex]\[
\text{Moles of } \text{MgCl}_2 = \text{Molarity of } \text{MgCl}_2 \times \text{Volume (in L)}
\][/tex]
[tex]\[
\text{Moles of } \text{MgCl}_2 = 0.0260 \, \text{mol/L} \times 1.55 \, \text{L} = 0.0403 \, \text{mol}
\][/tex]
4. Find the molar mass of [tex]\( \text{MgCl}_2 \)[/tex]: Calculate the molar mass by adding the atomic masses of magnesium and chlorine:
[tex]\[
\text{Molar mass of } \text{MgCl}_2 = 24.305 \, (\text{Mg}) + 2 \times 35.453 \, (\text{Cl}) = 95.211 \, \text{g/mol}
\][/tex]
5. Calculate the mass of [tex]\( \text{MgCl}_2 \)[/tex] needed: Use the formula for mass:
[tex]\[
\text{Mass of } \text{MgCl}_2 = \text{Moles of } \text{MgCl}_2 \times \text{Molar mass of } \text{MgCl}_2
\][/tex]
[tex]\[
\text{Mass of } \text{MgCl}_2 = 0.0403 \, \text{mol} \times 95.211 \, \text{g/mol} = 3.837 \, \text{g}
\][/tex]
Therefore, you need approximately 3.837 grams of [tex]\( \text{MgCl}_2 \)[/tex] to make a 1.55 L solution with a chloride ion concentration of 0.0520 M.
1. Understand the relationship in the chemical formula: Magnesium chloride ([tex]\( \text{MgCl}_2 \)[/tex]) dissociates in water to form one magnesium ion ([tex]\( \text{Mg}^{2+} \)[/tex]) and two chloride ions ([tex]\( \text{Cl}^- \)[/tex]). This means that each molecule of [tex]\( \text{MgCl}_2 \)[/tex] contributes two chloride ions to the solution.
2. Determine the molarity of [tex]\( \text{MgCl}_2 \)[/tex]: Since the solution requires a 0.0520 M concentration of chloride ions, and each mole of [tex]\( \text{MgCl}_2 \)[/tex] produces two moles of chloride ions, the concentration of [tex]\( \text{MgCl}_2 \)[/tex] needs to be half of the chloride ion concentration. Therefore, the molarity of [tex]\( \text{MgCl}_2 \)[/tex] is:
[tex]\[
\text{Molarity of } \text{MgCl}_2 = \frac{0.0520 \, \text{M}}{2} = 0.0260 \, \text{M}
\][/tex]
3. Calculate the moles of [tex]\( \text{MgCl}_2 \)[/tex] needed: Use the molarity formula, which is:
[tex]\[
\text{Moles of } \text{MgCl}_2 = \text{Molarity of } \text{MgCl}_2 \times \text{Volume (in L)}
\][/tex]
[tex]\[
\text{Moles of } \text{MgCl}_2 = 0.0260 \, \text{mol/L} \times 1.55 \, \text{L} = 0.0403 \, \text{mol}
\][/tex]
4. Find the molar mass of [tex]\( \text{MgCl}_2 \)[/tex]: Calculate the molar mass by adding the atomic masses of magnesium and chlorine:
[tex]\[
\text{Molar mass of } \text{MgCl}_2 = 24.305 \, (\text{Mg}) + 2 \times 35.453 \, (\text{Cl}) = 95.211 \, \text{g/mol}
\][/tex]
5. Calculate the mass of [tex]\( \text{MgCl}_2 \)[/tex] needed: Use the formula for mass:
[tex]\[
\text{Mass of } \text{MgCl}_2 = \text{Moles of } \text{MgCl}_2 \times \text{Molar mass of } \text{MgCl}_2
\][/tex]
[tex]\[
\text{Mass of } \text{MgCl}_2 = 0.0403 \, \text{mol} \times 95.211 \, \text{g/mol} = 3.837 \, \text{g}
\][/tex]
Therefore, you need approximately 3.837 grams of [tex]\( \text{MgCl}_2 \)[/tex] to make a 1.55 L solution with a chloride ion concentration of 0.0520 M.
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