What is the molar mass of iron ii sulfate feso4

Asked by admin @ 04/12/2021 in Chemistry viewed by 329 People

What mass of FeSO4^2- x 6H20 (Molar Mass=260g/mol) is required to produce 500 mL of a .10M iron (II) sulfate solution.

A.) 9g
B.) 13g
C.) 36g
D.) 72g

Answered by admin @ 04/12/2021

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: $M = \frac{n}{V (L)} = \frac{w}{m\times V(L)}$

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: $M = \frac{w}{m\times V(L)}$

⇒ $w = M\times m\times V(L)$

⇒ $w = (0.1 M)\times (260g/mol)\times (0.5L)$

⇒ mass of iron (II) sulfate taken: $w = 13 g$

Therefore, the mass of iron (II) sulfate taken for preparing the given solution is 13 g.

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