Answer:
The correct option is: B. 13g
Explanation:
Given: Molar mass of iron (II) sulfate: m = 260g/mol,
Molarity of iron (II) sulfate solution: M = 0.1 M,
Volume of iron (II) sulfate solution: V = 500 mL = 500 × 10⁻³ = 0.5 L (∵ 1L = 1000mL)
Mass of iron (II) sulfate taken: w = ? g
Molarity: [tex]M = \frac{n}{V (L)} = \frac{w}{m\times V(L)}[/tex]
Here, n- total number of moles of solute, w - given mass of solute, m- molar mass of solute, V- total volume of solution in L
∴ Molarity of iron (II) sulfate solution: [tex]M = \frac{w}{m\times V(L)}[/tex]
⇒ [tex]w = M\times m\times V(L)[/tex]
⇒ [tex]w = (0.1 M)\times (260g/mol)\times (0.5L) [/tex]
⇒ mass of iron (II) sulfate taken: [tex]w = 13 g[/tex]
Therefore, the mass of iron (II) sulfate taken for preparing the given solution is 13 g.
