Respuesta :
Answer:
0.64 M
Explanation:
Given:
Volume of iron(II) solution (V₁) = 25.0 mL = 0.025 L
Molarity of iron(II) solution (M₁) = ?
Number of moles of iron(II) solution (n₁) = ?
Volume of dichromate solution (V₂) = 18.0 mL = 0.018 L
Molarity of dichromate solution (M₂) = 0.145 M
Number of moles of dichromate solution (n₂) = ?
Molarity is equal to the ratio of moles and volume.
So, molarity of dichromate solution is given as:
[tex]M_2=\frac{n_2}{V_2}\\\\n_2=M_2\times V_2=0.145\times 0.018 = 2.61\times 10^{-3}\ mol[/tex]
Now, let us write the complete balanced reaction for the given situation.
So, the complete balanced equation is given below.
[tex]6Fe^{2+}(aq)+Cr_2O_7^{2-}(aq)+14H^+(aq)\to 6Fe^{3+}(aq)+2Cr^{3+}(aq)+7H_2O[/tex]
From the equation, it is clear that, 1 mole of dichromate is required for 6 moles of iron(II) solution.
So, using unitary method, we find the number of moles of iron(II) solution.
1 mole of dichromate = 6 moles of iron(II)
∴ n₂ moles of dichromate = 6n₂ moles of iron(II)
= [tex]6\times 2.61\times 10^{-3}=0.016\ mol\ Fe^{2+}[/tex]
So, 0.016 moles of iron(II) is needed. Therefore, [tex]n_1=0.016\ mol[/tex]
Now, molarity of iron(II) solution is given as:
Molarity = Moles ÷ Volume
[tex]M_1=\frac{n_1}{V_1}\\\\M_1=\frac{0.016\ mol}{0.025\ L}=0.64\ M[/tex]
Therefore, the molarity of the iron(II) solution is 0.64 M.
