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
The correct answer is: A
Explanation
Na2C2O4 + CaCl2 → CaC2O4 + 2NaClmolar 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