. Which volume of mercury (II) nitrate with a molar concentration of 0.25 mol/L should be taken as to prepare 100 ml of the solution with a titrimetric conversion factor corresponding to chloride ion=1,30010^-3g/ml

Question

. Which volume of mercury (II) nitrate with a molar concentration of 0.25 mol/L should be taken as to prepare 100 ml of the solution with a titrimetric conversion factor corresponding to chloride ion=1,30010^-3g/ml

Answer 1

To prepare 100 ml of a solution with a titrimetric conversion factor corresponding to chloride ion of 1.3 x 10^-3 g/ml, we need to calculate the amount of mercury(II) nitrate required.Given:- Molar concentration of mercury(II) nitrate = 0.25 mol/L- Volume of solution to be prepared = 100 mlTo calculate the amount of mercury(II) nitrate required, we can use the formula:Amount of mercury(II) nitrate = Molar concentration x VolumeSubstituting the given values:Amount of mercury(II) nitrate = 0.25 mol/L x 0.1 L = 0.025 molNow, we need to convert the amount of mercury(II) nitrate to grams using the molar mass of mercury(II) nitrate, which is 267.2 g/mol.Amount of mercury(II) nitrate in grams = 0.025 mol x 267.2 g/mol = 6.68 gTherefore, to prepare 100 ml of the solution with a titrimetric conversion factor corresponding to chloride ion of 1.3 x 10^-3 g/ml, you should take 6.68 grams of mercury(II) nitrate.

Вопрос

. Which volume of mercury (II) nitrate with a molar concentration of 0.25 mol/L should be taken as to prepare 100 ml of the solution with a titrimetric conversion factor corresponding to chloride ion=1,30010^-3g/ml

Решения

Ответ