A 0.600 g sample containing BaCl2 (molar mass 208 g mol^-1) is converted to BaCrO4, which is dissolved in acid (CrO4^2- becomes Cr2O7^2-). Excess KI is added and the iodine liberated needs 25.0 mL of 0.120 M Na2S2O3. The mass percent of BaCl2 is

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JEE Advanced levelSingle correct optionDifficulty 3 of 3eq092

A $0.600 g$ sample containing $BaCl X_{2}$ (molar mass $208 g mo l^{- 1}$ ) is converted to $BaCrO X_{4}$ , which is dissolved in acid ( $CrO X_{4} X^{2 -}$ becomes $Cr X_{2} O X_{7} X^{2 -}$ ). Excess $KI$ is added and the iodine liberated needs $25.0 mL$ of $0.120 M$ $Na X_{2} S X_{2} O X_{3}$ . The mass percent of $BaCl X_{2}$ is

A34.7%
B17.3%
C104.0%
D69.3%

Answer: A

Each Ba2+ gives one chromate, i.e. 0.5 dichromate, which liberates 1.5 I2 and needs 3 thiosulphate. mol thiosulphate = 0.0250 x 0.120 = 3.00e-3, so mol BaCl2 = 1.00e-03. Mass = 0.2080 g; percent = 0.2080/0.600 x 100 = 34.67%.

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A 0.600 g sample containing BaClX2​ (molar mass 208 gmol−1) is converted to BaCrOX4​, which is dissolved in acid (CrOX4​X2− becomes CrX2​OX7​X2−). Excess KI is added and the iodine liberated needs 25.0 mL of 0.120 M NaX2​SX2​OX3​. The mass percent of BaClX2​ is

Answer: A

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