The specific conductance of saturated solution of CaF2 is 3-86- 10-5 ohm cm-1 and that of water used for solution is 0-15- 10-5- The specific conductance of CaF2 alone is-

The correct option is A 3.71× 10^ 5 k (CaF_2 solution)=3.86 × 10^ 5 Ω^ 1cm^ 1 k (water solution)=0.15 × 10^ 5 Ω^ 1cm^ 1 k (CaF_2 alon ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Conductance and Conductivity

The specific ...

Question

The specific conductance of saturated solution of $C a F_{2}$ is $3.86 \times 10^{- 5}$ ohm $c m^{- 1}$ and that of water used for solution is $0.15 \times 10^{- 5}$ . The specific conductance of $C a F_{2}$ alone is:

3.71 \times 10^{- 5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4.01 \times 10^{- 5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.7 \times 10^{- 4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.86 \times 10^{- 4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $3.71 \times 10^{- 5}$
$k (C a F_{2} s o l u t i o n) = 3.86 \times 10^{- 5} Ω^{- 1} c m^{- 1}$
$k (w a t e r s o l u t i o n) = 0.15 \times 10^{- 5} Ω^{- 1} c m^{- 1}$
$k (C a F_{2} a l o n e) = ?$
$k (C a F_{2} a l o n e) = k (C a F_{2} s o l u t i o n) - k (W a t e r s o l u t i o n)$
$= (3.86 - 0.15) \times 10^{- 5} Ω^{- 1} c m^{- 1}$
$= 3.71 \times 10^{- 5} Ω^{- 1} c m^{- 1}$
The specific conductance of $C a F_{2}$ alone is $3.71 \times 10^{- 5} Ω^{- 1} c m^{- 1}$

Suggest Corrections

Similar questions

Q. The specific conductance of saturated solution of

Ca F_{2}

3.86 \times 10^{- 5} mho c m^{- 1}

and that water used for solution is

0.15 \times 10^{- 5}

. The specific conductance of

Ca F_{2}

alone is:

Q. At 291 K, the conductivity of saturated solution of

C a F_{2}

3.86 \times 10^{- 5} S c m^{- 1}

and that of water used for solution is

0.15 \times 10^{- 5} S c m^{- 1}

. The ionic conductance of

C a^{2 +}

and

F^{-}

at infinite dilution are 51.0 and 47.0

S c m^{2} m o l^{- 1}

respectively. The solubility of

C a F_{2}

in solution is

X \times 10^{- 2} g l i t r e^{- 1}

. Value of

X

is:

Q. Calculate the solubility product of the sparingly soluble salt

C a F_{2}

from the following data:
The molar ionic conductances (at infinite dilution) of

C a^{2 +}

and

F^{-}

ions are

104 \times 10^{- 4}

and

48 \times 10^{- 4} S m^{2} m o l^{- 1}

respectively. The specific conductance of the saturated solution of

C a F_{2}

at room temperature is

4.25 \times 10^{- 3} S m^{- 1}

and the specific conductance of water used for preparing the solution is

2 \times 10^{- 4} S m^{- 1}

Q. The specific conductance and the equivalent conductance of a saturated solution of

B a S O_{4}

8 \times 10^{- 5} o h m^{- 1} c m^{- 1}

and

8000 o h m^{- 1} c m^{2} m o l^{- 1}

respectively. The

K_{s p}

B a S O_{4}

is:

Q. The specific conductance of a saturated solution of AgCl is found to be

1.86 \times 10^{- 6} o h m^{- 1} c m^{- 1}

and that of water is

6 \times 10^{- 8} o h m^{- 1} c m^{- 1}

. The solubility of AgCl given that

λ_{A}^{0} g C l = (137.2 o h m^{- 1} c m^{- 1}

\times 10^{- 5} M

)

Join BYJU'S Learning Program

The specific conductance of saturated solution of CaF2 is 3.86×10−5 ohm cm−1 and that of water used for solution is 0.15×10−5. The specific conductance of CaF2 alone is:

The specific conductance of saturated solution of $C a F_{2}$ is $3.86 \times 10^{- 5}$ ohm $c m^{- 1}$ and that of water used for solution is $0.15 \times 10^{- 5}$ . The specific conductance of $C a F_{2}$ alone is: