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Chemistry Class 12th

When 200mL of 0.2 M acetic acid is shaken with 0.6g of wood charcoal, the final concentration of acetic acid after adsorption is 0.1M. The mass of acetic acid adsorbed per gram of carbon is ______________ g.

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Payal Gupta

Contributor-Level 10

Number of moles of CH₃COOH Adsorbed on 0.6 gm wood charcoal.

$= (0.2 * 200) * 1 0^{- 3} - (0.1 * 200) * 1 0^{- 3}$

= 20 * 10^-3 moles

Mass of CH₃COOH absorbed = 20 * 10^-3 * 60 = 1.2 gm

Mass of CH₃COOH adsorbed per gram = $\frac{1.2}{0.6} = 2 g$ of charcoal.

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Chemistry Chemistry Electrochemistry

The resistance of a conductivity cell containing 0.01 MKCl solution at 298K is $1750 Ω$ . If is 0.152 * 10^-3 cm^-1, then the cell constant of the conductivity cell _________ * 10^-3 cm^-1.

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Payal Gupta

Contributor-Level 10

$K = \frac{1}{R} * c e l l c o n s t a n t$

= 0.152 * 10^-3 * 1750

= 266 * 10^-3 cm^-1

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Chemistry Chemical Equilibrium

PCl₅ dissociates as

PCl₅(g) $?$ PCl₃ (g) + Cl₂(g)

5 moles of PCl₅ placed in a 200 litre vessel which contains 2 moles of N₂ and is maintained at 600K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl₅ is ____________ * 10^-3g. (nearest integer).

(Given R = 0.082 L atm K^-1 mol^-1; Assume ideal gas behavior)

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Payal Gupta

Contributor-Level 10

$\begin{array}{l} P C l_{5} (g) ? P C l_{3} (g) + C l_{2} (g) \\ t = 0 5 m o l e s 0 0 \\ \underline{- x m o l e} \underline{+ x m o l e} + \underline{x m o l e} \end{array}$

Eq.(5 – x) mole x molex mole

Total number of moles at equilibrium

$= (5 - x) + x + x + n_{N_{2}} = (7 + x)$

$K_{p} = \frac{(P_{P C l_{3}}) * {(P C l_{2})}_{C l}}{{(P_{P C l_{5}})}_{e q}} = \frac{(\frac{3}{10} * 2.46) * (\frac{3}{10} * 2.46)}{(\frac{2}{10} * 2.46)} = 1.107 a t m$

= 1107 * 103 atm

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Chemistry Chemistry Solutions

A Company dissolves 'x' amount of CO₂ at 298K in 1 litre of water to prepare soda water X = _________________ * 10^-3 g. (nearest integer)

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Payal Gupta

Contributor-Level 10

According to Henry's law

$P_{G a s} = K_{H} * x_{G a s}$

$= 0.835 = 1.67 * 1 0^{3} * [\frac{n_{C O_{2}}}{n_{C O_{2}} + 55.5}]$

As $n_{C O_{2}} < < 55.5$

$\Rightarrow n_{C O_{2}} = 27.78 * 1 0^{- 3} m o l e s$

Mass of CO₂ gas = $n_{C O_{2}} * (M M) C O_{2}$

= 1222 * 10^-3gm

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Chemistry Class 11th

At 300K, a sample of 3.0g of gas A occupies the same volume as 0.2g of hydrogen at 200K at the same pressure. The molar mass of gas A is _________g mol^-1, (nearest integer) Assume that the behavior of gases as ideal. (Given: The molar mass of hydrogen (H₂) gas is 2.0g mol^-1.)

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Payal Gupta

Contributor-Level 10

${P V)}_{H_{2}} = {(P V)}_{G a s}$

${(n R T)}_{H_{2}} = {(n R T)}_{G a s}$

$n_{H_{2}} * T_{H_{2}} = n_{G a s} * T_{G a s}$

$\Rightarrow \frac{0.2}{2} * 200 = \frac{3}{{(M M)}_{G a s}} * 300 \Rightarrow {(M M)}_{G a s} = 45 g m / m o l e$

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Chemistry Class 12th

The conversion of propan-1- ol to n- butylamine involves the sequential addition of reagents. The correct sequential order of reagent is

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alok kumar singh

Contributor-Level 10

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Chemistry Class 12th

Hex-4- ene- 2-ol on treatment with PCC gives 'A' on reaction with sodium hypolodite gives 'B', which on further heating with soda lime gives 'C'. The compound 'C' is

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