Suppose 24.9 grams of copper (II) nitrate is dissolved in 200. milliliters of a 0.70 moles/liter aqueous solution of sodium chromate.
Calculate the final molarity of nitrate anion in the solution.
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To calculate the final molarity of nitrate anion (NO3-) in the solution, we first need to find out the moles of copper (II) nitrate (Cu(NO3)2) and the volume of the resulting solution.
1. Calculate the moles of Cu(NO3)2:
Given that the mass of Cu(NO3)2 is 24.9 grams and its molar mass is 63.55 g/mol for copper, and 62.01 g/mol for nitrate, we can calculate the moles of Cu(NO3)2 using the equation:
moles = mass / molar mass
moles of Cu(NO3)2 = 24.9 g / (63.55 g/mol + (2 * 62.01 g/mol))
moles of Cu(NO3)2 = 24.9 g / 187.57 g/mol
moles of Cu(NO3)2 = 0.1327 mol
2. Calculate the Volume of the resulting solution:
Given that the volume of the sodium chromate solution is 200 mL, which is equivalent to 0.200 L.
3. Calculate the moles of nitrate anion (NO3-):
Since copper (II) nitrate dissociates into one Cu2+ ion and two NO3- ions, the number of moles of nitrate anion is given by:
moles of NO3- = 2 * moles of Cu(NO3)2
moles of NO3- = 2 * 0.1327 mol
moles of NO3- = 0.2654 mol
4. Calculate the final molarity of NO3-:
The final molarity of NO3- is given by:
Molarity = moles of NO3- / volume of solution (in liters)
Molarity = 0.2654 mol / 0.200 L
Molarity = 1.327 M
Therefore, the final molarity of nitrate anion (NO3-) in the solution is 1.327 M.