Question #8784a

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Answer 1 · 2017-12-11T12:52:20

This is a concentration and stoichiometry problem. We need,

$60mg * (g)/(10^3mg) * ("mol"NO_3^(-))/(62g) approx 9.7*10^-4"mol"$

of the nitrate ion. The way the compound you describe (relatively completely) dissociates in solution is,

$KNO_3(aq) to K^(+)(aq) + NO_3^(-)(aq)$

Hence,

$9.7*10^-4mol * (KNO_3)/(NO_3^(-)) * (101.1g)/(KNO_3) approx 9.8*10^-2g$

of $KNO_3$ per liter of distilled water are needed to arrive at the desired concentration.