Chemistry Test

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Chemistry Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

If mechanism of reaction is

Where k is rate constant then, what is $- \frac{d [A]}{d t}$ ?

A

\begin{equation}\frac{n K}{2}[A]\end{equation}

B
K[A]

C

\begin{equation}\frac{ n ( n +1)}{2} K [ A ]\end{equation}

D
Kⁿ[A]

Solution
Question 2
1 / -0

The ionic molar conductivities of $C _{2} O _{4}^{2 \Theta}, K ^{\oplus}$ and $Na ^{\ominus}$ ions are $x , y$ and $z S cm ^{2} mol ^{-1}$, respectively then value of $\wedge_{\text {eq }}^{\circ}$ of $( NaOOC - COOK )$ is

A
x + y + z

B
(x + y + z)/2

C
x – y – z

D
(x – y – z)/2

Solution
Question 3
1 / -0

0.5 molal solution acetic acid (M.W. = 60) in benzene (M.W. = 78) boils at 80.80 °C. The normal boiling point of benzene is 80.10 °C and Δ_vap H = 30.775 KJ/mol. Which of the following option is correct regarding percent of association of acetic acid in benzene.

A
53.2

B
0.532

C
93.6

D
0.936

Solution

Given to (boiling point of benzene) $=80.10^{\circ} C =353.1 K$ molality $( m )=0.5$ $\Delta_{ vap } H =30.775 KJ / mol$
$M W_{1}($ benzene $)=78$ $K _{ b }=\frac{ RT _{0}^{2} MW _{1}}{1000 \Delta_{ vap } H }=\frac{8.314 \times(353.1)^{2} \times 78}{1000 \times 30.775 \times 10^{3}}$
$=2.63$
$\Delta T_{b}=i K _{b} m \Rightarrow i =\frac{\Delta T _{b}}{ K _{b} \cdot m }=\frac{0.7}{2.65 \times 0.5}$
$i=0.532$
Considering association of acetic acid $2 CH _{3} COOH \rightleftharpoons\left( CH _{3} COOH \right)_{3}$
\begin{equation}\begin{array}{lll}
\text { Initially: } & 1 & 0 \\
\text { At equi: } & 1-\alpha & \lambda / 2
\end{array}\end{equation}Total number of moles $=1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$
$i=\frac{\text { Total number of moles }}{\text { Initial moles }}$
$0.532=1-\frac{\alpha}{2}$
$a=0.936$
$\%$ association $=93.6 \%$
Hence, option $C$ is correct.
Question 4
1 / -0

If NaCl is doped with 10^–2 mol% of SrCl₂, which of the following option shows concentration of cation vacancies?

A
10^–4 mol^–1

B
6.023 × 10¹⁹ mol^–1

C
6.023 × 10²¹ mol^–1

D
10⁶ mol^–1

Solution

Doping of $NaCl$ with $10^{-2} mol \% SrCl _{2}$ means that $100 mol$ of $NaCl$ are doped with $10^{-2} mol$ of $SrCl _{2}$
$100 mol$ of $NaCl$ doped with $=10^{-2} mol SrCl _{2}$
1 mol of NaCl doped with $=\frac{10^{-2}}{100}=10^{-4} mol SrCl _{2}$
Each $Sr ^{2}$ ion introduces one cation vacancy, therefore, cation vacancy: $10^{-4}$ mol/mol of NaCl $=10^{-4} \times 6.023 \times 10^{23} mol ^{-1}$
$=6.02 \times 10^{19} mol ^{-1}$ of $NaCl$
Hence, option $B$ is correct.
Question 5
1 / -0

Equilibrium constants of $T _{2} O$ (T or ${ }_{1}^{3} H$ is an isotope of ${ }_{1} H$ ) and $H _{2} O$ are different at $298 K$. At $298 K$ pure $T _{2} O$ has $p$ T (like $p _{ H }$ ) is $7.62 .$ The prof a solution prepared by adding $10 ml$ of $0.2 M$ TCI to $15 ml$ of $0.25 M$ NaOT is:

A
2 – log7

B
14 + log7

C
13.24 – log7

D
13.24 + log7

Solution

\begin{equation}\begin{array}{lcccc}
& & T Cl + NaOT & \rightarrow NaCl + T _{2} O \\
\text { Initial } & 2 & 3.75 & & \\
\text { Final } & 0 & 1.75 & 2 & 2
\end{array}\end{equation}milimoles of remaining $NaOT =1.75$ Volume $=10+25=25 m$
$\therefore[- OT ]=\frac{1.75}{25}=7 \times 10^{-2} moles$
$Por =-\log [- OT ]$
$Por =-\log \left(7 \times 10^{-2}\right)$
$p _{ OT }=2-\log 7$
since at $298 K$ for $T _{2} O , p _{ T }=7.62$
so, $p _{ T }+ p _{ OT }=7.62 \times 2$
$p _{ T }=15.24-(2-\log 7)$
$p _{ T }=13.24+\log 7$
Hence, option $D$ is correct.
Question 6
1 / -0

For a gas obeying the van der waals equation at critical temperature, which of the following is true.

A

\begin{equation}\operatorname{Both}\left(\frac{\partial P}{\partial V}\right)_{T} \text { and }\left(\frac{\partial^{2} P}{\partial V^{2}}\right)_{T} \text { are zero }\end{equation}

B

\begin{equation}\operatorname{Onl} y \cdot\left(\frac{\partial^{2} P}{\partial V^{2}}\right) \quad \text { is not zero }\end{equation}

C

\begin{equation}\left(\frac{\partial P}{\partial V}\right)_{T} \text { and }\left(\frac{\partial^{2} P}{\partial V^{2}}\right)_{T} \text { are not zero }\end{equation}

D

\begin{equation}\left(\frac{\partial P}{\partial V}\right)_{T} \text { is non zero but }\left(\frac{\partial^{2} P}{\partial V^{2}}\right)_{T} \text { is zero }\end{equation}

Solution

At critical point only one phase exits. Graph appears as given below

There is a stationary inflection point in PV Diagram (at a given critical temperature)
So, $\left(\frac{\partial P}{\partial V}\right)_{T}=0$
$\left(\frac{\partial^{2} P}{\partial V^{2}}\right)_{T}=0$
Hence, option (A) is correct.
Question 7
1 / -0

‘Blue baby’ syndrome is caused by –

A
Oxides of Sulphur

B
Nitrate in drinking water

C
Lead

D
Nitrogen Oxide

Solution

Blue baby syndrome is caused by nitrates in drinking water.

Hence, option (B) is correct.
Question 8
1 / -0

which of the following species undergo non–redox thermal decomposition reaction on heating

A
FeSO₄

B
SnSO₄

C
H₂C₂O₄

D
Na₂HPO₄

Solution

Non–redox decomposition implies no change in oxidation number

On taking a look at these options we observe only in option (D) there has been no change in oxidation number.

Hence, option (D) is correct
Question 9
1 / -0

In which of the following ‘meta form’ of ‘-ic acid’ is not possible.

A
B(OH)₃

B
H₂SO₄

C
H₃PO₄

D
None of these

Solution

For meta form of an acid to exist it must be capable of giving one H2O molecule and also contain at least one ‘H’ after donation of water.

A. $\mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{-\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{HBO}_{2}$ (Metaboric acid)
B. $\mathrm{H}_{2} \mathrm{SO}_{4} \stackrel{-\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \stackrel{S O_{3}}{\downarrow}$
an oxide not acid
$\mathrm{C} . \mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow-\mathrm{H}_{2} \mathrm{O} \rightarrow$ HPO $_{3}($ Metaphosphoric acid $)$
Hence, option (B) is correct.
Question 10
1 / -0

X + HNO₃→ Y + NO₂ + H₂O + S
Y + ammonium molybdate → yellow ppt.
Identity which of the following is (X):

A
As₂S₅

B
Sb₂S₅

C
PbS

D
NiS

Solution

As₂S₅ + HNO₃→ H₃AsO₄ + NO₂ + H₂O + S

H₃ASO₄ + (NH₄)₂MoO₄→(NH₄)₂4sO4 12MoO₃

Yellow ppt.

Hence option (A) correct.

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Chemistry Test - 14

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Chemistry Test - 14

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Question 1

\begin{equation}\frac{n K}{2}[A]\end{equation}

K[A]

\begin{equation}\frac{ n ( n +1)}{2} K [ A ]\end{equation}

Kn[A]

Solution

Question 2

x + y + z

(x + y + z)/2

x – y – z

(x – y – z)/2

Solution

Question 3

53.2

0.532

93.6

0.936

Solution

Question 4

10–4 mol–1

6.023 × 1019 mol–1

6.023 × 1021 mol–1

106 mol–1

Solution

Question 5

2 – log7

14 + log7

13.24 – log7

13.24 + log7

Solution

Question 6

\begin{equation}\operatorname{Both}\left(\frac{\partial P}{\partial V}\right)_{T} \text { and }\left(\frac{\partial^{2} P}{\partial V^{2}}\right)_{T} \text { are zero }\end{equation}

\begin{equation}\operatorname{Onl} y \cdot\left(\frac{\partial^{2} P}{\partial V^{2}}\right) \quad \text { is not zero }\end{equation}

\begin{equation}\left(\frac{\partial P}{\partial V}\right)_{T} \text { and }\left(\frac{\partial^{2} P}{\partial V^{2}}\right)_{T} \text { are not zero }\end{equation}

\begin{equation}\left(\frac{\partial P}{\partial V}\right)_{T} \text { is non zero but }\left(\frac{\partial^{2} P}{\partial V^{2}}\right)_{T} \text { is zero }\end{equation}

Solution

Question 7

Oxides of Sulphur

Nitrate in drinking water

Lead

Nitrogen Oxide

Solution

Question 8

FeSO4

SnSO4

H2C2O4

Na2HPO4

Solution

Question 9

B(OH)3

H2SO4

H3PO4

None of these

Solution

Question 10

As2S5

Sb2S5

PbS

NiS

Solution

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Kⁿ[A]

10^–4 mol^–1

6.023 × 10¹⁹ mol^–1

6.023 × 10²¹ mol^–1

10⁶ mol^–1

FeSO₄

SnSO₄

H₂C₂O₄

Na₂HPO₄

B(OH)₃

H₂SO₄

H₃PO₄

As₂S₅

Sb₂S₅