At higher altitude,the boiling point of water | vedclass

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

Question

(A) The elevation in boiling point is given by $\Delta T_b = i 	imes K_b 	imes m$. Here,$\Delta T_b = 100\,^oC - 95\,^oC = 5\,K$. For $NaCl$,the van

Vedclass · Accepted Answer

$296.05$

Vedclass · Answer

(A) The elevation in boiling point is given by $\Delta T_b = i 	imes K_b 	imes m$. Here,$\Delta T_b = 100\,^oC - 95\,^oC = 5\,K$. For $NaCl$,the van't Hoff factor $i = 1 + \alpha = 1 + 0.9 = 1.9$. The molality $m = \frac{w}{M 	imes W_{solvent(kg)}}$,where $w$ is the mass of $NaCl$ and $M = 58.5\,g/mol$. Substituting the values: $5 = 1.9 	imes 0.52 	imes \frac{w}{58.5 	imes 1}$. Solving for $w$: $w = \frac{5 	imes 58.5}{1.9 	imes 0.52} = \frac{292.5}{0.988} \approx 296.05\,g$.

At higher altitude,the boiling point of water is $95\,^oC$. The amount of $NaCl$ added to $1\, kg$ of water $(K_b = 0.52\, K\, kg\, mol^{-1})$ in order to raise the boiling point of the solution to $100\,^oC$ is ......... $g$. (Assuming $90\%$ ionization of $NaCl$)

Similar Questions

Which of the following will have the lowest vapor pressure?

$0.8 \ mL$ of acetic acid of density $1.06 \ g \ mL^{-1}$ when dissolved in $1 \ kg$ of water causes a depression in freezing point by $0.0325^{\circ} C$. The Van't Hoff factor is:
$(K_{f} \text{ of } H_2O = 1.86 \ K \ kg \ mol^{-1})$

Among the following $0.1 \ m$ aqueous solutions,which one will exhibit the lowest boiling point elevation,assuming complete ionization of the compound in solution?

Identify the compound amongst the following whose $0.1 \ M$ aqueous solution has the highest boiling point.

Which of the following is a non-electrolyte?

Vedclass Test Series

Exam Paper Generator

Online Exam Module