Respuesta :
Answer:
0.0398 mol·L⁻¹
Explanation:
You are diluting the solution with water, so you can use the dilution formula
[tex]c_{1}V_{1} = c_{2} V_{2}[/tex]
You can solve this equation to get
[tex]c_{2} = c_{1} \times \frac{V_{1}}{V_{2}}[/tex]
[tex]c_{1} = \text{0.157 mol/L}; V_{1} = \text{85.00 mL}[/tex]
[tex]c_{2} = ?; V_{2} = \text{85.00 mL + 250.0 mL} = \text{335.0 mL}[/tex]
[tex]c_{2} = \text{0.157 mol/L} \times \frac{\text{85.00 mL}}{ \text{335.0 mL}} = \text{0.0398 mol/L}[/tex]
The molarity of the resulting solution after adding the water is 0.0398 M.
The given parameters;
- volume of water, v = 250 mL
- volume of the original solution, v₁ = 85 mL
- initial concentration of the solution, c₁ = 0.157 M
The molarity of the resulting solution is calculated as follows;
[tex]C_1v_1 = C_2 v_2[/tex]
where;
- C₂ is the final volume of the solution = 250 ml + 85 ml = 335 ml
The molarity of the resulting solution is calculated as;
[tex]C_2 = \frac{C_1 v_1}{v_2} \\\\C_2 = \frac{0.157 \times 85}{335} \\\\C_2= 0.0398 \ M[/tex]
Thus, the molarity of the resulting solution after adding the water is 0.0398 M.
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