Home
Class 12
CHEMISTRY
What is meant by limiting molar conducti...

What is meant by limiting molar conductivity?

A

`Na^(+)`

B

`Mg^(2+)`

C

`K^(+)`

D

`Ca^(2+)`

Text Solution

Verified by Experts

The correct Answer is:
D

Ion `Na^(+)K^(+)Mg^(2+)Ca^(2+)`
50-70 73.50 106.0 119.0
`^^_(m)^(@)`
`(S cm^(2)"mol"^(-1))`
The size of hydrated `Na^(+)` ion is largest and that of `Ca^(2+)` is the smallest. Hence they follow the above order.
Promotional Banner

Similar Questions

Explore conceptually related problems

A saturated solution in AgA (K_(sp)=3xx10^(-14)) and AgB (K_(sp)=1xx10^(-14)) has conductivity of 375xx10^(-10) S cm^(-1) and limiting molar conductivity of Ag^(+) and A^(-) are 60S cm^(2) "mol"^(-1) and 80 S cm^(2) "mol"^(-1) respectively , then what will be the limiting molar conductivity of B^(-) (in S cm^(2) "mol"^(-1) )?

(a) What is limiting molar conductivity ? Why there is step rise in the molar conductivity of weak electrolyte on dilution ? (b) Calculate the emf of the following cell at 298 K : Mg(s) |Mg^(2+)(0.1 M)||Cu^(2+)(1.0 xx 10^(-3)M)|Cu(s) [Given : E_("cell")^(@) = 2.71 V]

Limiting molar conductivity of NaBr is

What is meant by humus?

(a) Apply Kohlrausch law of independent migration of ions, write the expression to determine the limiting molar conductivity of calcium chloride. (b) Given are the conductivity and molar conductivity of NaCI solutions at 298 K at different concentrations : Compare the variation of conductivity and molar conductivity of NaCI solutions on dilution. Give reason. (c) 0.1 M KCI solution offered a resistance of 100 ohms in conductivity cell at 298 K. If the cell constant of the cell is 1.29 cm^(-1) , calculate the molar conductivity of KCI solution.

The following curve is obtained when molar conductivity (wedge_(m)) is plotted against the square root of concentration, c^(1//2) for two electrolytes A and B (a) How do you account for the increase in the molar conductivity of the electrolyte A on dilution. (b) As seen from the graph, the value of limiting molar conductivity (wedge_(m)^(0)) for electrolyte B cannot be obtained graphically. How can this value be obtained?1

Calculate limiting molar conductivity of CaSO_(4) (limiting molar conductivity of Calcium and sulfate ions are 119.0 and 160.0 S cm ^(2)"mol"^(-1) respectively.

A weak monobasic acid is 5% dissociated in 0.01 mol dm^(-3) solution. The limiting molar conductivity at infinite dilution is 4.00xx10^(-2) ohm^(-1) m^(2) mol^(-1) . Calculate the conductivity of a 0.05 mol dm^(-3) solution of the acid.

What is meant by an octave.?