Question

The percentage weight of Zn in white vitriol $[ZnSO_4 . 7H_2O]$ is approximately equal to $(at. \, mass\, of \, Zn= 65, S=32, 0=16 \, and \, H= 1)$

• A. $33.65%$
• B. $32.56%$
• C. $23.65%$
• D. $22.65%$

Answer 1

The Correct Option is D

To calculate the percentage weight of Zinc (Zn) in white vitriol, which is the compound [ZnSO_4 . 7H_2O], we first need to determine the molar mass of the entire compound.

1. Calculate the molar mass of each component:
   • Zinc (Zn): Atomic mass = 65
   • Sulfur (S): Atomic mass = 32
   • Oxygen (O): Atomic mass = 16
   • Hydrogen (H): Atomic mass = 1
2. Calculate the molar mass of ZnSO_4:
   • Molar mass of ZnSO_4 = (Zn) + (S) + (4 × O) = 65 + 32 + (4 × 16) = 161
3. Calculate the molar mass of 7H_2O:
   • Molar mass of H_2O = (2 × H) + (O) = 2 × 1 + 16 = 18
   • Molar mass of 7H_2O = 7 × 18 = 126
4. Calculate the total molar mass of [ZnSO_4 . 7H_2O]:
   • Total molar mass = Molar mass of ZnSO_4 + Molar mass of 7H_2O = 161 + 126 = 287
5. Calculate the percentage weight of Zn in the compound:
   • Percentage weight of Zn = \(\frac{{\text{Molar mass of Zn}}}{{\text{Total molar mass of } [ZnSO_4 . 7H_2O]}} \times 100 = \frac{65}{287} \times 100 \approx 22.65\%\)

Therefore, the percentage weight of Zinc in white vitriol is approximately 22.65\%.