The molality of a solution,when 18 g of gluc | vedclass

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(C) Given: Mass of solute (glucose) $m_2 = 18 \ g$,Molar mass of glucose $M_2 = 180 \ g \ mol^{-1}$,Mass of solvent (water) $m_1 = 18 \ g$. Number of

Vedclass · Accepted Answer

$5.55$

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(C) Given: Mass of solute (glucose) $m_2 = 18 \ g$,Molar mass of glucose $M_2 = 180 \ g \ mol^{-1}$,Mass of solvent (water) $m_1 = 18 \ g$. Number of moles of solute $n_2 = \frac{m_2}{M_2} = \frac{18}{180} = 0.1 \ mol$. Mass of solvent in $kg$ is $m_1 = 18 \ g = 18 	imes 10^{-3} \ kg$. Molality $(m) = \frac{	ext{Number of moles of solute}}{	ext{Mass of solvent in } kg} = \frac{0.1}{18 	imes 10^{-3}} = \frac{0.1}{0.018} = 5.55 \ mol \ kg^{-1}$ or $5.55 \ m$.

The molality of a solution,when $18 \ g$ of glucose is added to $18 \ g$ of $H_2O$,is: (in $m$)

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