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3 years ago

5 drops of 0.15 M Ki added to40 drops of Na2S2O3What is the final concentration of ki?

Chemistry

1 answer:

rjkz [21]3 years ago

3 0

Answer:

$\boxed{\text{0.017 mol/L}}$

Explanation:

Na₂S₂O₃ solution does not react with KI (it reacts with I₂), so it is simply diluting the KI, and we can use the dilution formula.

$c_{1}V_{1} = c_{2}V_{2}$

Data:

c₁ = 0.15 mol·L⁻¹; V₁ = 5 drops

V(Na₂S₂O₃) = 40 drops

Calculations:

(a) Calculate the total volume

V₂ = 5 + 40 = 45 drops

(b) Calculate the concentration

0.15 × 5 = c₂ × 45

0.75 = 45c₂

$c_{2} = \dfrac{0.75 }{45} = \boxed{\textbf{0.017 mol/L}}$

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5 drops of 0.15 M Ki added to40 drops of Na2S2O3What is the final concentration of ki?​

5 drops of 0.15 M Ki added to40 drops of Na2S2O3What is the final concentration of ki?