A solution with a mass of 1.263 g containing an unknown amount of potassium ions was treated with excess sodium tetraphenylborate to ' precipitate 1.003 g of KB ${(C_{6} H_{5})}_{4}$ (M = 358.33. What is the mass percentage of potassium in the original solution?

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8.64%

13.8%

10.9%

9.16%

answer is A.

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Detailed Solution

$K^{+} + B {(C_{6} H_{5})}_{4}^{-} \to K_{B} {(C_{6} H_{5})}_{4}$

POAC on K

mole of K atom = mole of K atom in $KB {(C_{6} H_{5})}_{4}$

= mole of $KB {(C_{6} H_{5})}_{4}$

$\frac{x}{39} = \frac{1.003}{358.33}$

$x = 0.1092 gram$

% mass $= \frac{0.1092}{1.263} \times 100$

$= 8.64 %$

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