x/m =k.p1/n
log(x/m )=logk+ 1/ n logp
y=c+mx
m = 1/n and c= logk
2 R - OH + 2Na→ 2 RONa + H2↑2 moles of alcohol gives 1 mole of H2
which occupies 22.4L at 273K and 1 atm
∴ number of moles of alcohol = ( 2
moles of R - OH / 22.4 L of H2 ) × 560 mL
= 0.05 moles
∴ no. of moles = mass / molar mass
⇒ molar mass = 3.7/0.05 = 74 g mol−1
General
formula for R - OH Cn H2n+1 - OH
∴ n(12) + (2n+1) (1) + 16+1=74
14n = 74 – 18
14n = 56
∴ n = 56/14 = 4
The 2 alcohol which contains 4
carbon is CH2 CH(OH)CH2CH3
Ag2C2O4
↔ 2Ag+
+ C2O42-
[ Ag+ ] = 2.24 × 10-4
mol L-1
[ C2 O42-
] = { 2.24×10-4 }/2 = mol L-1
= 1.12×10-4 mol L-1
Ksp = [Ag+]2[C2O42-]
= (2.24×10-4 mol-4 L-1
)2 (1.12×10-4 mol L-1)
= 5.619×10-12mol3
L-3
1F = 96500 C = 1 mole of e− = 6.023 ×1023
e−
∴ 9650 C = [ 6.22 ×1023 / 96500 ] × 9650 = 6.022×1022
= 6.022×1022
The incorrect statement is option
(b)
Physisorption is an exothermic
process. Hence increase in temperature decreases the physisorption.