A volume of mL of chlorine gas (at STP) is reacted with excess solution. The amount of liberated can — Redox Reactions and Volumetric Analysis Chemistry Question
Question
A volume of $10$ mL of chlorine gas (at STP) is reacted with excess $KI$ solution. The amount of $I_2$ liberated can be titrated with $x$ mL of $0.1 M$ $Na_2S_2O_3$ solution. $x$ is:
Answer: A
💡 Solution & Explanation
Moles $Cl_2 = 10 / 22400$. Eq $Cl_2$ ($n=2$) = Eq $Na_2S_2O_3$ ($n=1$). $2 \times (10/22400) = 0.1 \times x/1000 \Rightarrow x = 20000 / 2240 = 8.92$ mL. [1, 5]
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