Chemistry, asked by omansg, 9 months ago

please post solutions photo.
and your paytm no.
I will give 10 rupee for correct solution.
pls units bhi dekh lena and writing bekar mat batana​

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Answered by MrIntrovert27
2

Answer:

1. ∧(m) at infinity  = number of moles of cations × λ(cation) + number of moles of anions × λ(anion)

So,

∧(m) at infinity = 1 * 300

and We Know

∧(m) at infinity = ∧(e) at infinity × n-factor

300 = ∧(e) at infinity × 3

∧(e) at infinity = 100 S cm^2 eq^(-1)

2. We Can Write Al2(SO4)3 as 2Al^(+3) + 3SO4^(-2)

∧(m) at infinity  = number of moles of cations × λ(cation) + number of moles of anions × λ(anion)

∧(m) at infinity = 2*300 + 3*125

∧(m) at infinity = 975

∧(m) at infinity = ∧(e) at infinity × n-factor

975  = ∧(e) at infinity × 6

∧(e) at infinity  = 162.5 S cm^2 eq^(-1)

3. We Can Write (NH4)2SO4 as NH4^(+1) + SO4(-2)

∧(m) at infinity  = number of moles of cations × λ(cation) + number of moles of anions × λ(anion)

∧(m) at infinity = 1*200 + 2*125

∧(m) at infinity = 450 S cm^2 mol^(-1)

Or hn bhai meri hisse ke 10 Rupey ki chips kha lena xD

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