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Class 10 Class 12

Electrochemistry

Calculate the molar conductance at infinite dilution of ethanoic acid from the following data:
Λ°_m (HCl) = 425.9 s cm² mol^–1Λ°_m (CH₃COONa) = 91.0s cm² mol^–1Λ°_m (NaCl) = 126.4 s cm² mol^–1

Λ°_m (CH₃COOH) = λ°_H++ λ°_CH₃coo^–= Λ°_m(HCl) = λ°_m (CH₃COONa) – λ°_NaCl= 425.9 + 91.0 –126.4
= 390.5 s cm² mol^–1.

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The specific conductance of a saturated solution of AgCl in water is 1.826 x 10^–6ohm^–1cm^–1 at 25°C. Calculate its solubility in water at 25°C. [Given Λ^∞_m (Ag⁺) = 61.92 ohm^–1 cm² mol^–1and Λ^∞_m (CI^– ) = 76.34 ohm^–1 cm² mol^–1]

Λ^∞_m (AgCl) = λ^∞_m_m (Ag⁺) + λ^∞_m (Cl^–)

= 61.92 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} + 76.34 {ohm}^{- 1} {cm}^{2} {mol}^{- 1} = 138.26 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}

K = 1.826 \times 10^{6} {ohm}^{- 1} {cm}^{2}

Solubility (in mol L^-1) =

\frac{K \times 1000 {cm}^{3} L^{- 1}}{A_{m}^{\infty}}

= \frac{(1.826 \times 10^{- 6} {ohm}^{- 1}) \times (1000 {cm}^{3} L^{- 1})}{(138.26 {ohm}^{- 1} {cm}^{2} {mol}^{- 1})} = 1.32 \times 10^{- 5} mol L^{- 1} .

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How long will it take to deposit electrolytically 127 gm of copper on the cathode of a copper voltameter by a current of 50A? Given ECE of copper = 0.0003294 g/c.

We have given that

m = 127 gm,
I = 50 A
Z = 0.0003294 g/c
t = ?
By Faraday's first law,
m = ZIt

or $t = \frac{m}{ZI} = \frac{127}{0.0003294 \times 50}$

t = 77118 sec.

138 Views

The molar conductivity of KCl solutions at different concentration at 298 K are given below:

C/mol L^–1

Λ/s cm² mol^–1

0.000198

0.000309

0.000521

0.000989

148.61

148.29

147.81

147.09

Show that a plot between A and C^1/2 is a straight line. Determine the values of Λ° and Λ for KCl.

Taking the square root of concentration, we have

C^1/2/(mol L^–1)^1/2

A/s cm² mol^–1

0.01407

0.01758

0.02283

0.03145

148.61

148.29

147.81

147.09

A plot of A(l /-axis) and c^1/2 (x-axis) is shown in the fig. It is nearly a straight line. From the intercept (C^1/2 = 0) we find that Λ° = 149.8 s cm² mol^–1 and Slope = A = 87.46s cm²mol^–1/(mol / L^–1)^1/2.

355 Views

An electric current of 0.5 ampere was passed through acidulated water for one hour. Calculate the volume of hydrogen at STP produced. 1 coulomb of electricity deposits 0.00001 gm of hydrogen.

We have given that
current = 0.5 ampere
weight of H₂ (W) = ?

t = 1 hour = 3600 sec,
Z = 0.00001.

W = Z x c x t

= 0.00001 x 0.5 x 3600 g = 0.018 g
2g or H₂ at STP occupies = 22.4 litres
$∴$ 0.018 g of H₂ at STP occupies

$\frac{22.4}{2} \times 0.018 = 0.2016 litres .$
Hence, Volume of H₂ produced at STP = 0.2016 litres.

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Electrochemistry

An electric current of 0.5 ampere was passed through acidulated water for one hour. Calculate the volume of hydrogen at STP produced. 1 coulomb of electricity deposits 0.00001 gm of hydrogen.