Question

Question #8784a

Answer 1

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.