How many grams of KI would you need to prepare 0.50 g of 7.0% KI (m/m) solution?

fomula; ( x = mutiplacation btw)

% m/m = m(solute) / m(solution) x 100%

0.50g x 70% = m/m x 100%

0.5g x 70% / 100% = 3.5 g KI

(both % cancle each other)

Explanation:

heads up might be wrong i did it myself

i need mutiple choices

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