Calculate the molar mass of a nonvolatile sol | vedclass

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

Question

(B) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \cdot m$,where $m$ is the molality of the solution.Molality $m = \frac{W_2 	imes

Vedclass · Accepted Answer

$175 \ g \ mol^{-1}$

Vedclass · Answer

(B) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \cdot m$,where $m$ is the molality of the solution.Molality $m = \frac{W_2 	imes 1000}{M_2 	imes W_1}$,where $W_2$ is the mass of solute,$M_2$ is the molar mass of solute,and $W_1$ is the mass of solvent.Substituting the values: $\Delta T_{b} = 0.01 \ K$,$K_{b} = 0.50 \ K \ kg \ mol^{-1}$,$W_2 = 0.35 \ g$,and $W_1 = 100 \ g$.$0.01 = 0.50 	imes \frac{0.35 	imes 1000}{M_2 	imes 100}$.$M_2 = \frac{0.50 	imes 0.35 	imes 1000}{0.01 	imes 100} = \frac{175}{1} = 175 \ g \ mol^{-1}$.

Calculate the molar mass of a nonvolatile solute if a solution containing $0.35 \ g$ of solute in $100 \ g$ of water has a boiling point elevation of $0.01 \ K$ $\left[K_{b}=0.50 \ K \ kg \ mol^{-1}\right]$.

Similar Questions

$0.1 \ molal$ 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 $^{\circ} C$)?

$x \ g$ of urea (molar mass $60 \ g \ mol^{-1}$) is completely dissolved in $y \ g$ of pure water and the solution boiled at $373.202 \ K$. If the boiling point of pure water at $1.013 \ bar$ is $373.15 \ K$,then $x:y$ is $(K_b(H_2O) = 0.52 \ K \ kg \ mol^{-1})$

The elevation in boiling point of a solution containing $1.8 \ g$ of glucose in $100 \ g$ of solvent is $0.1 \ ^\circ C$. Find the molal elevation constant of the liquid.

Which of the following aqueous solutions will exhibit the highest boiling point?

When common salt is dissolved in water,

Vedclass Test Series

Exam Paper Generator

Online Exam Module