Question

Calculate molality of each ion present in aqueous solution of 2m nh4cl

Answer 1

The given molality of ammonium chloride $NH_{4}Cl$ is 2 m

Ammonium chloride gives the following ions in solution,

$NH_{4}Cl (aq) ---> NH_{4}^{+} (aq) + Cl^{-} (aq)$

Each mole of ammonium chloride gives one mole ammonium ion($NH_{4}^{+}$) and one mole chloride($Cl^{-}$) ion.

The molality of ammonium ion:

$\frac{2 mol NH_{4}Cl }{1 kg water} *\frac{1 mol NH_{4}^{+}  }{1 mol NH_{4}Cl } =2 m NH_{4}^{+}$

The molality of chloride ion:

$\frac{2mNH_{4}Cl }{1 kg water} * \frac{1 mol Cl^{-} }{1 mol NH_{4}Cl } = 2 m Cl^{-}$