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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$0.18$

Vedclass · Answer

(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

$45 \ g$ of ethylene glycol $(C_{2}H_{6}O_{2})$ is mixed with $600 \ g$ of water. Calculate $(a)$ the freezing point depression and $(b)$ the freezing point of the solution.

Camphor is often used in molecular mass determination because

$x$ moles of $CO(NH_2)_2$ are present in $1200 \ g$ of water. If the freezing point of the solution is $-4.02 \ ^oC$,calculate the value of $x$. Given $k_f \ (H_2O) = 1.86 \ K \ kg \ mol^{-1}$.

Calculate the cryoscopic constant $(K_f)$ of a solvent if the depression in freezing point of a $0.18 \ m$ solution of a non-volatile solute is $0.2 \ K$.

When an aqueous solution freezes at its freezing point,which components are in equilibrium?

Vedclass Test Series

Exam Paper Generator

Online Exam Module