Question

$8.50\, gm$ of $NH_3$ is present in $250\, mL$ volume. Its active mass is:

• A. $1.0 ML^{-1}$
• B. $0.5 ML^{-1}$
• C. $1.5 ML^{-1}$
• D. $2.0\, ML^{-1}$

Answer 1

Answer: (D)

Active mass = Number of moles of $NH _{3}$ present in $1 L$ solution

$\therefore $ Active mass $=\frac{8.5\, g / 17\, g\, mol ^{-1}}{250\, mL / 1000\, mL ^{-1}}$

$=2\, ML ^{-1}$

(concentra in molarity or mole per litre)