Answer :
The concentration of [tex]\rm Mg^2^+[/tex] ions in the 500.0 mL of diluted solution is approximately 0.4808 M.
We will calculate the initial concentration of [tex]\rm Mg^2^+[/tex]ions in the original solution.
Given:
Total volume of original solution ([tex]\rm V_1[/tex]) = 1000.0 mL
Mass of [tex]\rm Mg(NO_3)_2[/tex] (m) = 37.1 g
The molar mass of [tex]\rm Mg(NO_3)_2[/tex] = 24.31 g/mol (Mg) + 2 * 14.01 g/mol (N) + 6 * 16.00 g/mol (O)
= 24.31 g/mol + 28.02 g/mol + 96.00 g/mol
= 148.33 g/mol
Number of moles of [tex]\rm Mg(NO_3)_2[/tex] = Mass / Molar mass
Number of moles of [tex]\rm Mg(NO_3)_2[/tex] = 37.1 g / 148.33 g/mol
Number of moles of [tex]\rm Mg(NO_3)_2[/tex] ≈ 0.2500 mol
Let us now determine the initial [tex]\rm Mg^2^+[/tex]ion concentration ([tex]\rm C_1[/tex]) in the original solution. [tex]\rm Mg(NO_3)_2[/tex] produces one [tex]\rm Mg^2^+[/tex] ion for each formula unit.
Initial concentration ([tex]\rm C_1[/tex]) = Number of moles / Volume
Initial concentration ([tex]\rm C_1[/tex]) = 0.2500 mol / 1000.0 mL
Initial concentration ([tex]\rm C_1[/tex]) = 0.2500 mol / 1.000 L
Initial concentration ([tex]\rm C_1[/tex]) = 0.2500 M (moles per liter)
Calculating the concentration of [tex]\rm Mg^2^+[/tex] ions in the diluted solution.
Given for the diluted solution:
Volume of diluted solution ([tex]\rm V_2[/tex]) = 500.0 mL
Volume of removed solution ([tex]\rm V_r_e_m_o_v_e_d[/tex]) = 20.0 mL
Volume after dilution = [tex]\rm V_2 + V_r_e_m_o_v_e_d[/tex]
Volume after dilution = 500.0 mL + 20.0 mL
Volume after dilution = 520.0 mL = 0.520 L
The number of moles of [tex]\rm Mg^2^+[/tex]ions in the removed solution remains the same as the initial solution.
Final concentration ([tex]\rm C_2[/tex]) = Number of moles / Volume
Final concentration ([tex]\rm C_2[/tex]) = 0.2500 mol / 0.520 L
Final concentration ([tex]\rm C_2[/tex]) ≈ 0.4808 M (moles per liter)
Therefore, the concentration of [tex]\rm Mg^2^+[/tex] ions in the 500.0 mL of diluted solution is approximately 0.4808 M.
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Final answer:
To find the concentration of Mg2+ ions in the 500.0 mL solution, first calculate the number of moles of Mg(NO3)2 in the original solution. Then use this information to determine the concentration of Mg2+ ions in the 500.0 mL solution.
Explanation:
To find the concentration of Mg2+ ions in the 500.0 mL solution, we need to first determine the number of moles of Mg2+ ions in the original 1000.0 mL solution. The molar mass of Mg(NO3)2 can be calculated using the periodic table: Mg = 24.31 g/mol, N = 14.01 g/mol, O = 16.00 g/mol. Therefore, the molar mass of Mg(NO3)2 is 24.31 + 2(14.01) + 6(16.00) = 148.33 g/mol. Next, we can calculate the number of moles of Mg(NO3)2 in the original solution:
37.1 g Mg(NO3)2 ÷ 148.33 g/mol = 0.25 mol Mg(NO3)2
Now that we know the number of moles of Mg(NO3)2 in the original solution, we can use this information to calculate the concentration of Mg2+ ions in the 500.0 mL solution:
0.25 mol Mg(NO3)2 ÷ 1.000 L = 0.25 M
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