# Question 8784a

Dec 11, 2017

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,

$K N {O}_{3} \left(a q\right) \to {K}^{+} \left(a q\right) + N {O}_{3}^{-} \left(a q\right)$

Hence,

$9.7 \cdot {10}^{-} 4 m o l \cdot \frac{K N {O}_{3}}{N {O}_{3}^{-}} \cdot \frac{101.1 g}{K N {O}_{3}} \approx 9.8 \cdot {10}^{-} 2 g$

of $K N {O}_{3}$ per liter of distilled water are needed to arrive at the desired concentration.