The active mass of 64 g of HI in a 2 L flas | vedclass

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(D) The active mass is defined as the molar concentration in $mol \ L^{-1}$.First,calculate the molar mass of $HI$: $1 + 127 = 128 \ g \ mol^{-1}$.Nex

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(D) The active mass is defined as the molar concentration in $mol \ L^{-1}$.First,calculate the molar mass of $HI$: $1 + 127 = 128 \ g \ mol^{-1}$.Next,calculate the number of moles of $HI$: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{64 \ g}{128 \ g \ mol^{-1}} = 0.5 \ mol$.Finally,calculate the active mass (concentration): $[HI] = \frac{n}{V} = \frac{0.5 \ mol}{2 \ L} = 0.25 \ mol \ L^{-1}$.

The active mass of $64 \ g$ of $HI$ in a $2 \ L$ flask is:

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