JEE MAIN - Chemistry (2004 - No. 38)
The limiting molar conductivities Λ° for NaCl, KBr and KCl are 126, 152 and 150 S cm2 mol-1 respectively. The Λ° for NaBr is
128 S cm2 mol-1
278 S cm2 mol-1
176 S cm2 mol-1
302 S cm2 mol-1
Explanation
$${A^ \circ }Nacl = {\lambda ^ \circ }N{a^ + } + \lambda C{l^ - }\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
$${A^ \circ }KBr = {\lambda ^ \circ }{K^ + } + \lambda Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
$${A^ \circ }KCl = {\lambda ^ \circ }{K^ + } + \lambda C{l^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$
operating $$(i)+(ii)-(iii)$$
$${A^ \circ }NaBr = {\lambda ^ \circ }N{a^ + } + {\lambda ^ \circ }B{r^ - }$$
$$ = 126 + 152 - 150$$
$$ = 128\,\,\,S\,c{m^2}\,mo{l^{ - 1}}$$
$${A^ \circ }KBr = {\lambda ^ \circ }{K^ + } + \lambda Br\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
$${A^ \circ }KCl = {\lambda ^ \circ }{K^ + } + \lambda C{l^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$$
operating $$(i)+(ii)-(iii)$$
$${A^ \circ }NaBr = {\lambda ^ \circ }N{a^ + } + {\lambda ^ \circ }B{r^ - }$$
$$ = 126 + 152 - 150$$
$$ = 128\,\,\,S\,c{m^2}\,mo{l^{ - 1}}$$
Comments (0)
