3. You will need to make ~0.1 M solutions of one (or more) of the following salts. For each compound, determine (or look up) its molecular weight, and determine the mass necessary to make 100.0 mL of 0.100 M solution. Pay attention to significant figures! Sodium acetate anhydrous Sodium acetate trihydrate Iron(III) chloride hexahydrate g

Question

contestada

Respuesta :

kobenhavn kobenhavn · Answer 1 · 2021-04-13T03:31:39+02:00

Answer: The mass of sodium acetate anhydrous required is 0.820 g

The mass of sodium acetate trihydrate required is 1.36 g

The mass of Iron (III) chloride required is 2.70 g

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n = moles of solute

= volume of solution in ml

1. moles of =

Now put all the given values in the formula of molarity, we get

[tex]0.100=\frac{x\times 1000}{82.03\times 100.0}[/tex]

[tex]x=0.820g[/tex]

Therefore, the mass of sodium acetate anhydrous required is 0.820 g

2. moles of [tex]CH_3COONa.3H_2O[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{136.08g/mol}[/tex]

Now put all the given values in the formula of molarity, we get

[tex]0.100=\frac{x\times 1000}{136.08\times 100.0}[/tex]

[tex]x=1.36[/tex]

Therefore, the mass of sodium acetate trihydrate required is 1.36 g

3. moles of [tex]FeCl_3.6H_2O[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{g/mol}[/tex]

Now put all the given values in the formula of molarity, we get

[tex]0.100=\frac{x\times 1000}{270.33\times 100.0}[/tex]

[tex]x=2.70g[/tex]

Therefore, the mass of Iron (III) chloride required is 2.70 g