If K_b denotes the molal elevation constant o | vedclass

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

Question

(D) The molality $m$ of the solution is calculated as follows:Mass of glucose $(W_2)$ $= 36 \ g$.Molar mass of glucose $(M_2)$ $= 180 \ g \ mol^{-1}$.

Vedclass · Accepted Answer

$(100 + \frac{2 \ K_b}{10})^{\circ} C$

Vedclass · Answer

(D) The molality $m$ of the solution is calculated as follows:Mass of glucose $(W_2)$ $= 36 \ g$.Molar mass of glucose $(M_2)$ $= 180 \ g \ mol^{-1}$.Volume of solution $= 1 \ dm^3 = 1 \ L$.Assuming the density of the solution is approximately $1 \ g \ mL^{-1}$,the mass of the solvent (water) $W_1$ is $1000 \ g = 1 \ kg$.Molality $m = \frac{W_2}{M_2 	imes W_1 (	ext{in } kg)} = \frac{36}{180 	imes 1} = 0.2 \ mol \ kg^{-1} = \frac{2}{10} \ mol \ kg^{-1}$.The elevation in boiling point is given by $\Delta T_b = K_b 	imes m = K_b 	imes \frac{2}{10} = \frac{2 \ K_b}{10}$.The boiling point of water is $100^{\circ} C$.Therefore,the boiling point of the solution $T_b = 100 + \Delta T_b = (100 + \frac{2 \ K_b}{10})^{\circ} C$.

If $K_b$ denotes the molal elevation constant of water,then the boiling point of an aqueous solution containing $36 \ g$ of glucose (molar mass $= 180 \ g \ mol^{-1}$) per $dm^3$ is:

Similar Questions

$A$ solution of $5.6 \ g$ nonvolatile solute in $50 \ g$ solvent has an elevation in boiling point of $1.75 \ K$. What is the molar mass of the solute $\left(K_{b} = 3 \ K \ kg \ mol^{-1}\right)$?

$0.1 \ molal$ aqueous solution of glucose boils at $100.16^{\circ} C$. What is the boiling point of a $0.5 \ molal$ aqueous solution of glucose (in $^{\circ} C$)?

What will be the boiling point of $1 \ m$ urea solution in $K$ unit? $(K_b = 0.52 \ K \ kg \ mol^{-1})$

The molal boiling point constant for water is $0.513 \, ^oC \, kg \, mol^{-1}$. When $0.1 \, mol$ of sugar is dissolved in $200 \, mL$ of water,the solution boils under a pressure of one atmosphere at ......... $^oC$.

By studying the elevation in boiling point of an aqueous solution of glucose,the value of $K_b$ for water is found to be $0.51 \ K \ kg \ mol^{-1}$. What value of $K_b$ can be expected from an aqueous solution of $K_4[Fe(CN)_6]$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module