The ^⁰ values of KNO₃ and LiNO₃ are 145.0 and 110.1 Scm² mol⁻¹ respectively. The λ⁰ value
for K⁺ ion is 73.5 S cm² mol⁻¹. Calculate λ⁰ (Li⁺).
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Answers
Answer:
in this question 1M solution of NaNO
3
,
By definition if Molarity we know that it is the no. of moles present in per litre of solution. = 1M molar solution has 1 mole of NaNO
3
for 1000ml of solution.
Therefore, mass of solute is mass of 1 mole of NaNO
3
= 23+14+48 = 85 gm.
volume of solution = 1000 ml as per definition.
Now,Given that density is 1.25 g/cm3
Therefore, mass = 1.25×1000 (m = density X volume)
=> mass = 1250 gm (solution mass)
Mass of solvent = mass of solution - mass of solute
= 1250 - 85= 1165gm.
Molarity is the no. of mole of solute per kg of solvent
=> m =
1165
1
×1000 =
1165
1000
=0.85m
Answer:
∵ LiCl.3NH
3
(s)⇌LiCl.3NH
3
(s)+2NH
3
(g)
[K
p
=9atm
2
]
LiCl.3NH
3
(s)+2NH
3
⇌LiCl.3NH
3
(s)
[K
p
1
=
9
1
atm
−2
]
0.1 (a + 0.2) Initial moles
0 a 0.1 Final Moles at equilibrium
∴ Initial moles of NH
3
should be (a+0.02) to bring in completion of reaction.
At
equilibrium K
p
1
=
(P
NH
3
′
)
2
1
or
9
1
=
(P
NH
3
′
)
2
1
∴P
NH
3
′
= 3 atm
∵ PV=nRT
∴ 3×5=n×0.0821×313
∴ n = 0.5837 i.e. a = 0.5837
∴ Initial moles of NH
3
= a + 0.2 = 0.5837 + 0.2=0.7837 moles