$0.600 g$ of pure $CaCO X_{3}$ (molar mass $100 g mo l^{- 1}$ ) is treated with $100 mL$ of $0.200 M$ $HCl$ . The excess acid is back-titrated with $0.100 M$ $NaOH$ . The volume of $NaOH$ needed is

A140 mL
B80 mL
C200 mL
D120 mL

Answer: B

mol CaCO3 = 6.00e-3, consuming 2 x 6.00e-3 = 1.20e-02 mol HCl. HCl taken = 0.100 x 0.200 = 2.00e-02 mol. Excess = 8.00e-03 mol, needing 8.00e-03/0.100 = 80 mL of NaOH.

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0.600 g of pure CaCOX3​ (molar mass 100 gmol−1) is treated with 100 mL of 0.200 M HCl. The excess acid is back-titrated with 0.100 M NaOH. The volume of NaOH needed is

Answer: B

More questions on Back titration and hardness of water

$0.600 g$ of pure $CaCO X_{3}$ (molar mass $100 g mo l^{- 1}$ ) is treated with $100 mL$ of $0.200 M$ $HCl$ . The excess acid is back-titrated with $0.100 M$ $NaOH$ . The volume of $NaOH$ needed is