Chemistry - Online Test
The complex is [M(H2O)5
Cl] Cl2
1000 ml of 1M solution of the
complex gives 2 moles of Cl− ions
1000 ml of 0.01M solution of the
complex will give
[100 ml x 0.01M x 2Cl-] /
[1000 ml x 1M] = 0.002 moles of C− ions
number of A ions = (Nc/8) = (8/8)=1
number of B ions = (Nf/2) = (6/2) =3
Following solutions were prepared by mixing different volumes of NaOH of
HCl different concentrations.
i. 60 mL (M/10) HCl + 40mL (M/10) NaOH
ii. 55 mL (M/10) HCl + 45 mL (M/10) NaOH
iii. 75 mL (M/5) HCl + 25mL (M/5) NaOH
iv. 100 mL (M/10) HCl + 100 mL (M/10) NaOH
iii) 75 ml M/5 HCl + 25ml M/5
NaOH
No of moles of HCl = 0.2×75×10-3
= 15 × 10-3
No of moles of NaOH = 0.2 × 25 ×
10-3 = 5 × 10-3
No of moles of HCl after mixing =
15 × 10-3 - 5 × 10-3
= 10 × 10-3
concentration
of HCl = No of moles of HCl / Vol in litre
= 10 ×10−3 / 100 ×10−3 = 0.1M
for (iii) solution, pH of 0.1M
HCl = -log10(0.1)
= 1.
Consider the following half cell reactions:
Mn2+ + 2e− → Mn Eº = -1.18V
Mn2+ → Mn3+
+ e-
Eº = -1.51V
The E for the reaction 3Mn2+ → Mn+2Mn3+ , and the possibility of the
forward reaction are respectively.
Mn2+ + 2e− → Mn (Eºred) = -1.18V
2[Mn2+ → Mn3+
+ e- ](Eºox) = −1.51V
3Mn2+ → Mn + 2Mn3+ Eºcell?
Eºcell = ( Eºox )+ ( Eºred )
= −1.51 −1.18 and non spontaneous
= −2.69V
Since Eo is –ve ∆G is +ve and the given forward cell reaction is non
– spontaneous.
B(OH)3 + H2O ↔
[B (OH)4 + H+
