Question

The Active Mass Of 64 G Of Hi In A 2 Lit Flask Would Be _____.A)2 B)1 C)5 D)0 .25

Answer 1

☯ GiveN :

Volume = 2L

Given Mass of HI = 64g

$\rule{200}{1}$

☯ To FinD :

We have to find the active mass of HI

$\rule{200}{1}$

☯ SolutioN :

For doing this question firstly we will calculate the molecular mass of HI.

$\sf{\dashrightarrow Molecular \: mass \: of \: HI = Molecular \: mass of \: H + Molecular \: mass \: of I} \\ \\ \sf{\dashrightarrow Molecular \: mass \: of \: HI = 1.0079 + 126.9045} \\ \\ \sf{\dashrightarrow Molecular \: mass \: of \: HI = 127.9124} \\ \\ \sf{\therefore \: Molecular \: mass \: of \: HI \: is \: 128 \: g \: mol^{-1}}$

$\rule{150}{2}$

Now,

$\small{\star{\boxed{\sf{Active \: mass = \dfrac{Given \: mass}{Molecular \: of \: HI \times Volume} }}}}$

★ Putting Values ★

$\sf{\dashrightarrow Active \: mass = \dfrac{64}{128 \times 2}} \\ \\ \sf{\dashrightarrow Active \: mass = \dfrac{64}{256}} \\ \\ \sf{\dashrightarrow Active \: mass = 0.25} \\ \\ \Large{\implies{\boxed{\boxed{\sf{Active \: mass = 0.25}}}}} \\ \\ \sf{\therefore \: Option \: D \: is \: correct.}$

Answer 2

Answer:

GiveN :

Volume = 2L

Given Mass of HI = 64g

☯ To FinD :

We have to find the active mass of HI

☯ SolutioN :

For doing this question firstly we will calculate the molecular mass of HI.

\begin{lgathered}\sf{\dashrightarrow Molecular \: mass \: of \: HI = Molecular \: mass of \: H + Molecular \: mass \: of I} \\ \\ \sf{\dashrightarrow Molecular \: mass \: of \: HI = 1.0079 + 126.9045} \\ \\ \sf{\dashrightarrow Molecular \: mass \: of \: HI = 127.9124} \\ \\ \sf{\therefore \: Molecular \: mass \: of \: HI \: is \: 128 \: g \: mol^{-1}}\end{lgathered}

⇢MolecularmassofHI=MolecularmassofH+MolecularmassofI

⇢MolecularmassofHI=1.0079+126.9045

⇢MolecularmassofHI=127.9124

∴MolecularmassofHIis128gmol

−1

Now,

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