a) Take 5-6 drops of a solution of any chromium (III) salt and add dropwise a diluted solution of sodium hydroxide until the precipitate of chromium hydroxide is formed. Divide the precipitate of Cr(OH)_(3) into two parts and dissolve them in NaOH and HCl respectively. __ ICl respective nm

To determine the number of moles of chromium (III) salt in the solution, we need to follow these steps:<br /><br />1. Calculate the molar mass of chromium (III) hydroxide, $Cr(OH)_3$.<br />2. Determine the mass of the precipitate formed.<br />3. Calculate the number of moles of $Cr(OH)_3$ using its molar mass.<br />4. Since each mole of $Cr(OH)_3$ contains one mole of chromium (III), the number of moles of chromium (III) salt will be the same as the number of moles of $Cr(OH)_3$.<br /><br />Let's assume that the mass of the precipitate formed is $m$ grams.<br /><br />1. Molar mass of $Cr(OH)_3$:<br /> - Chromium (Cr): 52.00 g/mol<br /> - Oxygen (O): 16.00 g/mol<br /> - Hydrogen (H): 1.01 g/mol<br /><br /> Molar mass of $Cr(OH)_3$ = 52.00 + (3 × 16.00) + (3 × 1.01) = 103.00 g/mol<br /><br />2. Number of moles of $Cr(OH)_3$:<br /> Number of moles = mass / molar mass<br /> Number of moles = $m$ grams / 103.00 g/mol<br /><br />3. Number of moles of chromium (III) salt:<br /> Since each mole of $Cr(OH)_3$ contains one mole of chromium (III), the number of moles of chromium (III) salt will be the same as the number of moles of $Cr(OH)_3$.<br /><br />Therefore, the number of moles of chromium (III) salt in the solution is $m$ grams / 103.00 g/mol.