8.50 gm of NH3 is present in 250 mL volume. Its active mass is:

Question

8.50 gm of NH3 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

Tardigrade Admin · Accepted Answer

Active mass = Number of moles of NH 3 present in 1 L solution ∴ Active mass =(8.5 g / 17 g mol -1/250 mL / 1000 mL -1) =2 ML -1 (concentra in molarity or mole per litre)

Q. $8.50 g m$ of $N H_{3}$ is present in $250 m L$ volume. Its active mass is:

$1.0 M L^{- 1}$

$0.5 M L^{- 1}$

$1.5 M L^{- 1}$

$2.0 M L^{- 1}$

Active mass = Number of moles of $N H_{3}$ present in $1 L$ solution

$∴$ Active mass $= \frac{8.5 g /17 g m o l ^{- 1}}{250 m L /1000 m L ^{- 1}}$

$= 2 M L^{- 1}$

(concentra in molarity or mole per litre)

Q. 8.50gm of NH3​ is present in 250mL volume. Its active mass is:

1.0ML−1

0.5ML−1

1.5ML−1

2.0ML−1