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10.1 g of KNO(3) is dissolved in 500 mL ...

`10.1 g` of `KNO_(3)` is dissolved in `500 mL` of `H_(2) O`. Mass of `Ba(NO_(3))_(2)` that should be added to this solution of get a molality `(m)` of 0.3 with respect to `NO_(3)^(ɵ)` ion is
`(Mw of KNO_(3) = 101 g mol^(-1), Mw of Ba(No_(3))_(2) = 261 g mol^(-1))`
a. `~~ 1.3 g` b. `~~ 13 g` c.`~~ 6.5 g` d. `~~ 65 g`

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To solve the problem, we need to find the mass of barium nitrate (Ba(NO₃)₂) that should be added to a solution of potassium nitrate (KNO₃) to achieve a specific molality of nitrate ions (NO₃⁻). Here’s a step-by-step breakdown of the solution: ### Step 1: Calculate the moles of KNO₃ First, we need to calculate the number of moles of KNO₃ in the solution. \[ \text{Moles of KNO}_3 = \frac{\text{mass of KNO}_3}{\text{molar mass of KNO}_3} \] ...
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