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What is the mole fraction of a solution prepared by dissolving 50.0 g of potassium permanganate [tex]$KMnO_4$[/tex] in 75.0 g of water?

A) 0.076
B) 0.071
C) 7.06
D) 0.55

Sagot :

GinnyAnswer
To determine the mole fraction of a solution prepared by dissolving 50.0 grams of potassium permanganate ([tex]$KMnO_4$[/tex]) in 75.0 grams of water ([tex]$H_2O$[/tex]), follow these steps:

### Step 1: Calculate the Molar Mass of [tex]$KMnO_4$[/tex]

First, determine the molar mass of [tex]$KMnO_4$[/tex]. The atomic masses are:
- Potassium (K): 39.1 g/mol
- Manganese (Mn): 54.9 g/mol
- Oxygen (O): 16.0 g/mol

The molar mass of [tex]$KMnO_4$[/tex] is calculated as:
[tex]\[ 39.1 + 54.9 + (16 \times 4) = 39.1 + 54.9 + 64 = 158.0 \text{ g/mol} \][/tex]

### Step 2: Calculate the Number of Moles of [tex]$KMnO_4$[/tex]

Using the mass of [tex]$KMnO_4$[/tex] (50.0 grams) and its molar mass (158.0 g/mol), calculate the moles of [tex]$KMnO_4$[/tex]:
[tex]\[ \text{Moles of } KMnO_4 = \frac{50.0 \text{ g}}{158.0 \text{ g/mol}} = 0.316455696 \text{ moles} \][/tex]

### Step 3: Calculate the Molar Mass of Water ([tex]$H_2O$[/tex])

The molar mass of water ([tex]$H_2O$[/tex]) is calculated using the atomic masses of hydrogen (H) and oxygen (O):
- Hydrogen (H): 1.01 g/mol each
- Oxygen (O): 16.0 g/mol

The molar mass of [tex]$H_2O$[/tex] is:
[tex]\[ (2 \times 1.01) + 16 = 2.02 + 16 = 18.02 \text{ g/mol} \][/tex]

### Step 4: Calculate the Number of Moles of Water

Using the mass of water (75.0 grams) and its molar mass (18.02 g/mol), calculate the moles of water:
[tex]\[ \text{Moles of water} = \frac{75.0 \text{ g}}{18.02 \text{ g/mol}} = 4.162042175 \text{ moles} \][/tex]

### Step 5: Calculate the Total Number of Moles

Sum the moles of [tex]$KMnO_4$[/tex] and water to get the total moles:
[tex]\[ \text{Total moles} = \text{Moles of } KMnO_4 + \text{Moles of water} = 0.316455696 + 4.162042175 = 4.478497871 \text{ moles} \][/tex]

### Step 6: Calculate the Mole Fraction of [tex]$KMnO_4$[/tex]

The mole fraction of [tex]$KMnO_4$[/tex] is the ratio of the moles of [tex]$KMnO_4$[/tex] to the total moles:
[tex]\[ \text{Mole fraction of } KMnO_4 = \frac{\text{Moles of } KMnO_4}{\text{Total moles}} = \frac{0.316455696}{4.478497871} = 0.070661124 \][/tex]

### Conclusion

The mole fraction of the solution prepared by dissolving 50.0 g of potassium permanganate in 75.0 g of water is approximately 0.071. Therefore, the correct answer is:

[tex]\[ \boxed{0.071} \][/tex]

So, the correct choice is:
[tex]\[ \text{B) 0.071} \][/tex]
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