. 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

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.<br /><br />Given:<br />- Molar concentration of mercury(II) nitrate = 0.25 mol/L<br />- Volume of solution to be prepared = 100 ml<br /><br />To calculate the amount of mercury(II) nitrate required, we can use the formula:<br /><br />Amount of mercury(II) nitrate = Molar concentration x Volume<br /><br />Substituting the given values:<br /><br />Amount of mercury(II) nitrate = 0.25 mol/L x 0.1 L = 0.025 mol<br /><br />Now, 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.<br /><br />Amount of mercury(II) nitrate in grams = 0.025 mol x 267.2 g/mol = 6.68 g<br /><br />Therefore, 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.