Question

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

Answer 1

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

Step-by-step explanation:

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

`Molarity=(n* 1000)/(V_s)`

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

`0.100=(x* 1000)/(82.03* 100.0)`

`x=0.820g`

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

2. moles of
`CH_3COONa.3H_2O` =
`\frac{\text {given mass}}{\text {Molar mass}}=(xg)/(136.08g/mol)`

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

`0.100=(x* 1000)/(136.08* 100.0)`

`x=1.36`

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

3. moles of
`FeCl_3.6H_2O` =
`\frac{\text {given mass}}{\text {Molar mass}}=(xg)/(g/mol)`

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

`0.100=(x* 1000)/(270.33* 100.0)`

`x=2.70g`

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