A compound H_2X with molar weight of 80 g m | vedclass

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(C) Given: Molarity $(M)$ = $3.2 \ mol \ L^{-1}$.Consider $1 \ L$ of the solution.Number of moles of solute $(n)$ = $3.2 \ mol$.Mass of solute = $n

Vedclass · Accepted Answer

$8$

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(C) Given: Molarity $(M)$ = $3.2 \ mol \ L^{-1}$.Consider $1 \ L$ of the solution.Number of moles of solute $(n)$ = $3.2 \ mol$.Mass of solute = $n 	imes 	ext{Molar mass} = 3.2 \ mol 	imes 80 \ g \ mol^{-1} = 256 \ g$.Mass of solution = $	ext{Density of solution} 	imes 	ext{Volume}$. Since the density of the solvent is $0.4 \ g \ mL^{-1}$ and we assume no volume change,we consider the solvent volume as $1 \ L$ $(1000 \ mL)$.Mass of solvent = $	ext{Density} 	imes 	ext{Volume} = 0.4 \ g \ mL^{-1} 	imes 1000 \ mL = 400 \ g = 0.4 \ kg$.Molality $(m)$ = $\frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{3.2 \ mol}{0.4 \ kg} = 8 \ m$.

$A$ compound $H_2X$ with molar weight of $80 \ g \ mol^{-1}$ is dissolved in a solvent having density of $0.4 \ g \ mL^{-1}$. Assuming no change in volume upon dissolution,the molality of a $3.2 \ M$ solution is:

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