Chemistry Test

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Chemistry Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

The solubility s of potassium nitrate is 140 g per 100 g of water at 70.9^◦C, which decreases to 63.6 g per 100 g of water at 39.9^◦C. The enthalpy of crystallization, ∆H_cryst

A
22.1 kJ mol-1

B
22.1 Jmol-1

C
22.1 kcalmol-1

D
22.1 calmol-1

Solution

For convenience, we will call the higher temperature T2 and the lower temperature T1.
(1) The van’t Hoff isochore, Equation is written in terms of a ratio, so
we do not need the absolute values.
$\ln \left(\frac{K_{\mathrm{s}(2)}}{K_{\mathrm{s}(1)}}\right)=\frac{-\Delta H_{\mathrm{(cryst})}^{\bullet}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)$
(2) We convert the two temperatures to kelvin, for the van’t Hoff isochore
requires thermodynamic temperatures, so T2 = 343.9 K and T1 = 312.0 K.
(2) Putting the values in the equation
$\ln \left(\frac{140}{63.6}\right)=-\frac{-\Delta H^{\circ}}{8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1}} \times\left(\frac{1}{343.9 \mathrm{K}}-\frac{1}{312.0 \mathrm{K}}\right)$ $\ln (2.20)=\frac{-\Delta H_{(\operatorname{cryst})}^{\bullet}}{8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1}} \times\left(-2.973 \times 10^{-4} \mathrm{K}^{-1}\right)$
$\ln 2.20=0.7889$
We then divide both sides by ‘2.973 × 10-4 K-1’ so:$$\frac{0.7889}{2.973 \times 10^{-4} \mathrm{K}^{-1}}=\frac{\Delta H_{(\mathrm{cryst})}^{\bullet}}{8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1}}$$The term on the left cquals $2.654 \times 10^{3} \mathrm{K}$. Multiplying both sides by $R$ then yiclds:$$\Delta H_{(\mathrm{cryst})}^{\bullet}=2.654 \times 10^{3} \mathrm{K} \times 8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$$so$
\Delta H_{(c) \Delta \theta}^{\bullet}=22.1 \mathrm{kJ} \mathrm{mol}^{-1}
$
Question 2
1 / -0

The Clark cell Zn|Zn²⁺// Hg₂SO₄|Hg is often employed as a standard cell since its emf is known exactly as a function of temperature. The cell emf is 1.423 V at 298 K and its temperature coefficient of voltage is -1.2 × 10^-4 V K^-1. What is ∆S_cell at 298 K?

A
-23.2 JK^-1mol^-1

B
23.2 JK^-1mol^-1

C
-2.32 JK^-1mol^-1

D
2.32 JK^-1mol^-1

Solution
Question 3
1 / -0

An element has a bcc structure with a cell edge of 288 pm. The density of the element is 7.2 g cm^–3. Number of atoms present in 208 g of the element ?

A
2.416 × 10²³ atoms

B
24.16 × 10²³ atoms

C
241.6 × 10²³ atoms

D
0.2416 × 10²³ atoms

Solution

= 12.08 × 1023 unit cells

Since each bcc cubic unit contains 2 atoms, therefore the total number of atoms in 208 g

= 2 (atoms/unit cell) × 12.08 × 1023 unit cells

= 24.16 × 1023 atoms
Question 4
1 / -0

Which of the following is correct?

A
Change in Gibbs free energy is minimum at equilibrium

B
Gibbs free energy is minimum at equilibrium

C
Change in Gibbs free energy with respect to reaction progress is always positive

D
Change in Gibbs free energy with respect to reaction progress is always negative.

Solution

At equilibrium ∆G=0 which implies G will be minimum.
Graphs of Gibbs free energy with Reaction progress or extent of reaction are given below

As we can see in graph (a) the reaction is having ∆G <0 in which case equilibrium is achieved at K>1
In graph (b) the reaction is having ∆G>0 in which case equilibrium is achieved at K<1
In graph (c) the reaction is having ∆G=0 in which case equilibrium is achieved at K=1
Question 5
1 / -0

Ratio of Boyle temperature and critical temperature of gas is:

A
8/27

B
27/8

C
1/2

D
2

Solution

Boyle Temperature T_b= a/R_b

Critical Temperature T_c= 8a/27R_b

Dividing both

T_b/T_c =27/8
Question 6
1 / -0

Among the following, the surfactant that will form micelles in aqueous solution at lowest molar concentration at ambient condition is

A
CH₃(CH₂)₁₇N(CH₃)₃Cl

B
CH₃(CH₂)₁₉OSO₃K

C
CH₃(CH₂)₆COONa

D
CH₃(CH₂)₁₁N(CH₃) ₃I

Solution

Critical concentration for micelle formation decreases as the molecular weight of hydrocarbon chain of surfactant grows because in this case true solubility diminishes and the tendency of surfactant molecules to associate increases.
Question 7
1 / -0

Which of the following is wrong regarding titration of CH₃COOH with NaOH?

A
There is buffer formation.

B
Equivalent point is at pH >7

C
Phenolphthalein can be used an indicator

D
If we take weak acid then effect of weak acid is neutralized by weak base and pH at equivalent point is 7.

Solution

a) CH₃COOH + NaOH → CH₃COONa + H₂

T=0 C₁ X

Since we are titrating CH₃COOH with NaOH, concentration of NaOH will be less than CH₃COOH. Suppose x amount of NaOH is used.

T=t C₁-X X

As a result CH₃COONa will be formed in X amount. And (C1-X) amount of CH₃COOH and X amount of CH₃COONa will form buffer..

b) Since, strong base is used equivalent point will appear after 7.

c) In this case phenolphthalein is used as indicator.

d) Weak acid and weak base titration cannot be clearly predicted.
Question 8
1 / -0

Arrhenius relation is described as K= A e^-Ea/RT which of the following statement is correct regarding activation energy

A
Lower activation energy reactions are more susceptible to temperature change

B
Higher activation energy reactions are more susceptible to temperature change

C
Activation energy is thermo dynamical concept while rate is kinetic concept both are fundamentally different

D
Activation energy is of reaction is independent from temperature.

Solution

$\mathrm{K}=A e^{-E a / R T}$

Differentiating with respect to T $\frac{\partial K}{\partial T}=-K \frac{E_{a}}{R T^{2}}$

$\frac{\partial K / K}{\partial T}=-\frac{E_{a}}{R T^{2}}$

Here we can see that on increasing temperature those reaction having high value of E_a will be more susceptible to change in rate.
Question 9
1 / -0

Which of the following is reducing sugar?

A
Glucose

B
Fructose

C
Galactose

D
All of these

Solution

A reducing sugar is one which is capable of acting as reducing agent.

All monosaccharaides are reducing sugars because all monosaccharaides have an aldehyde group (if they are aldoses) or can tautomerize in solution to form an aldehyde group (if they are ketoses) and can themselves get oxidized hence can act as reducing agent.
Question 10
1 / -0

The tautomer of the shown compound is:

A
Aromatic

B
Anti-Aromatic

C
Aliphatic

D
Aldehyde

Solution

The tautomer of the shown compound is

We see that there is 4 pi-electrons and by Huckel rule, we know that the compound is Anti-aromatic.

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Chemistry Test - 19

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Chemistry Test - 19

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Question 1

22.1 kJ mol-1

22.1 Jmol-1

22.1 kcalmol-1

22.1 calmol-1

Solution

Question 2

-23.2 JK-1mol-1

23.2 JK-1mol-1

-2.32 JK-1mol-1

2.32 JK-1mol-1

Solution

Question 3

2.416 × 1023 atoms

24.16 × 1023 atoms

241.6 × 1023 atoms

0.2416 × 1023 atoms

Solution

Question 4

Change in Gibbs free energy is minimum at equilibrium

Gibbs free energy is minimum at equilibrium

Change in Gibbs free energy with respect to reaction progress is always positive

Change in Gibbs free energy with respect to reaction progress is always negative.

Solution

Question 5

8/27

27/8

1/2

2

Solution

Question 6

CH3(CH2)17N(CH3)3Cl

CH3(CH2)19OSO3K

CH3(CH2)6COONa

CH3(CH2)11N(CH3) 3I

Solution

Question 7

There is buffer formation.

Equivalent point is at pH >7

Phenolphthalein can be used an indicator

If we take weak acid then effect of weak acid is neutralized by weak base and pH at equivalent point is 7.

Solution

Question 8

Lower activation energy reactions are more susceptible to temperature change

Higher activation energy reactions are more susceptible to temperature change

Activation energy is thermo dynamical concept while rate is kinetic concept both are fundamentally different

Activation energy is of reaction is independent from temperature.

Solution

Question 9

Glucose

Fructose

Galactose

All of these

Solution

Question 10

Aromatic

Anti-Aromatic

Aliphatic

Aldehyde

Solution

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-23.2 JK^-1mol^-1

23.2 JK^-1mol^-1

-2.32 JK^-1mol^-1

2.32 JK^-1mol^-1

2.416 × 10²³ atoms

24.16 × 10²³ atoms

241.6 × 10²³ atoms

0.2416 × 10²³ atoms

CH₃(CH₂)₁₇N(CH₃)₃Cl

CH₃(CH₂)₁₉OSO₃K

CH₃(CH₂)₆COONa

CH₃(CH₂)₁₁N(CH₃) ₃I