Na₂C₂O₄ + CaCl₂ → CaC₂O₄ + 2NaCl. Given a solution of 1.9g of sodium oxalate in 50g of water at room temperature, calculate the minimum volume of 0.1 M calcium oxalate using the above equation?

1.40 χ 10² dm³

14.0 χ 10² cm³

1.40 χ 10^-2 dm^-2

14.0 χ 10^-2 cm³

Correct answer is A

Na₂C₂O₄ + CaCl₂ → CaC₂O₄ + 2NaCl

molar mass of Na₂C₂O₄ = 134

molarity in 1000g H₂O = (1.9)/(134) = 0.014 m

but in 502 g mole = (0.014)/(50) χ 1000 = 0.28 m

M₁V₁ = M₂V₂
V₂ = (M₁V₁)/(M₂) = (0.28 χ 50)/(0.1) = 140
= 1.40 χ 10² dm³

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Na₂C₂O₄ + CaCl₂ → CaC₂O₄ + 2NaCl. Given a solution of 1.9g of sodium oxalate in 50g of water at room temperature, calculate the minimum volume of 0.1 M calcium oxalate using the above equation?

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Na₂C₂O₄ + CaCl₂ → CaC₂O₄ + 2NaCl. Given a solution of 1.9g of sodium oxalate in 50g of water at room temperature, calculate the minimum volume of 0.1 M calcium oxalate using the above equation?