How many grams of glucose must be added to 1 | vedclass

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(D) The depression of freezing point is given by the formula: $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Rearranging this,w

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$0.18$

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(D) The depression of freezing point is given by the formula: $\Delta T_f = K_f 	imes m$,where $m$ is the molality of the solution.Rearranging this,we get $\frac{\Delta T_f}{K_f} = m$.Given that $\frac{\Delta T_f}{K_f} = \frac{1}{1000} = 0.001 \ mol/kg$.Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.Since the density of water is $1 \ g/mL$,$1 \ L$ of water is $1 \ kg$.Therefore,the number of moles of glucose required is $0.001 \ mol$.The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$.Mass of glucose = $	ext{moles} 	imes 	ext{molar mass} = 0.001 \ mol 	imes 180 \ g/mol = 0.18 \ g$.

How many grams of glucose must be added to $1 \ L$ of water such that the value of $\frac{\Delta T_f}{K_f}$ becomes $\frac{1}{1000}$?

Similar Questions

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