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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Ionic Equilibrium

NTA Abhyas 2020 · Chemistry · Ionic Equilibrium

The Solubility of A B _ 2 is 0.05g per 100mL at 25^ o C. Calculate...

The Solubility of AB2A{B}_{2} is 0.05g\text{0}\text{.05}\text{g} per 100mL\text{100}\text{mL} at {\text{25}^{o} \text{C}\text{.} Calculate Ksp{K}_{\text{sp}} of AB2A{B}_{2} at {\text{25}^{o} \text{C?} [Atomic mass of A=20amu,\text{A}\text{=}\text{20}\text{amu,} atomic mass of B=40amu\text{B}\text{=}\text{40}\text{amu}]

Options

  1. A10−31{0}^{- 3}
  2. B5×10−75 \times 1{0}^{- 7}
  3. C10−61{0}^{- 6}
  4. D5×10−35 \times 1{0}^{- 3}

Correct answer

B. 5×10−75 \times 1{0}^{- 7}

Step-by-step solution

=0.05g/100mL\text{=}\text{0}\text{.05}\text{g/100}\text{mL}
=0.5g/L\text{=}\text{0}\text{.5}\text{g/L}
=0.5100=5×10−3mol/L= \dfrac{0.5}{100} = 5 \times {10}^{- 3} \text{mol/L}
Ksp=4s3=4×(5×10−3)3=5×10−7{K}_{sp} = 4s^{3} = 4 \times {\left(5 \times 1{0}^{- 3}\right)}^{3} = 5 \times 1{0}^{- 7}
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