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from the following molar conductivities at infinite dilution,^m at infinte for Al2(SO4)3 = 858 cm2mol-1.^m at infinite for NH4OH = 238.3 S cm2 mol-1^m at infinity for (NH4)2SO4= 238.4 S cm2 mol-1.. Calculate ^m infinite for Al(OH)3

Shriya Mehrotra , 10 Years ago
Grade 12
anser 2 Answers
Ramreddy

Last Activity: 10 Years ago

^m (Al2(SO4)3) = 2*^Al+3 + 3 ^SO4 2-
^m (NH4OH) = ^ NH4+ + ^ OH-
^m ((NH4)2SO4) = 2 * ^ NH4+ + ^SO42-
^m (Al(OH)3) = ^Al+3 + 3 OH-
2 * ^m (Al(OH)3) = ^m (Al2(SO4)3) + 6* ^m (NH4OH) – 2 * ^m (Al(OH)3)

This is the formula you can get your answer from above one. just substitute.......

ankit singh

Last Activity: 4 Years ago

OH) – 2 * ^m (Al(OH)3)4) + 6* ^m (NH3)4(SO2 + 3 OH-2 * ^m (Al(OH)3) = ^m (Al+3) = ^Al3^m (Al(OH)2-4 + ^SO+4) = 2 * ^ NH4SO2 + ^ OH-^m ((NH4)+4^m (NH4OH) = ^ NH2- 4 + 3 ^SO+3^m (Al2(SO4)3) = 2*^Al

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