Consider the galvanic cell, Pt (s) `|H_(2) (1"bar")|` HCl (aq) (1M) `|Cl_(2) ("1 bar")|` Pt(s). After running the cell for sometime, the concentration of the electrolyte is automatically raised to 3M HCl. Molar conductivity of the 3M HCl is about `240S cm^(2) mol^(-1)` and limiting molar conductivity of HCl is about `420S cm^(2) mol^(-1)`. If `K_(b)` of water is 0.52K kg `mol^(-1)`, calculate the boiling point of the electrolyte at the end of the experiment

Options

A. 375.6K
B. 376.3K
C. 378.1K
D. 380.3K

Answer

A

Explanation

Given, `Lamda_(m)= 240S cm^(2) mol^(-1)`
`Lamda_(m)^(oo) =240S cm^(2) mol^(-1) " " K_(b)= 0.52K kg mol^(-1)`
Degree of ionisation, `alpha= (Lamda_(m))/(Lamda_(m)^(oo))=(240 Scm^(2) mol^(-1))/(420 S cm^(2) mol^(-1))=0.57`
van.t Hoff factor `i=1+ (n-1) alpha` [for ionisation]
`=1 + (2-1)0.57` [for HCl, n=2]
=1.57
Elevation in boiling, `Delta T_(b)= iK_(b) xx` molality (m) `=1.57 xx 0.52K kg mol^(-1) xx 3 mol kg^(-1)` =2.45K
Since, water boils at 373.15K at 1 bar pressure, therefore the boiling point of solution will be
`T_(b)= T_(b)^(@) + Delta T_(b)= 373.15+2.45`
=375.6K