Some basic concepts OF Chemistry
189 Questions
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Q176
Errorless
JEE Section
MSQ
1 mol $BaF_{2} + 2mol H_{2}SO_{4}$ → resulting mixture will be neutralised by
A.
1 mol of KOH
B.
2 mol of $Ca(OH)_{2}$
C.
4 mol of KOH
D.
2 mol of KOH
Q177
Errorless
JEE Section
MSQ
A sample of $H_{2}O_{2}$ solution labelled as "28 volume" has density of 26.5 g/L. Mark the correct option(s) representing concentration of same solution in other units
A.
$M_{H_{2}O_{2}} = 2.5$
B.
$\% \frac{w}{v} = 17$
C.
Mole fraction of $H_{2}O_{2}=0.2$
D.
$m_{H_{2}O_{2}} = 13.88$
Q178
Errorless
JEE Section
MSQ
Two bulbs A and B contains $16g O_{2}$ and $16g O_{3}$ ,
respectively. Which of the statements are true
A.
Both bulbs contain same number of atoms
B.
Both bulbs contain different number of atoms
C.
Both bulbs contain same number of molecules
D.
Bulb A contains $N_{A} / 2$ molecules while bulb B contains $N_{A} / 3$ molecules ( $N_{A} = \text{Avogadro's number}$ )
Q179
Errorless
JEE Section
MCQ
Statement 1 : If 30 mL of $H_{2}$ and 20 mL of $O_{2}$ react to form water, 5 mL of $H_{2}$ is left at the end of the reaction.
Statement 2 : $H_{2}$ is the limiting reagent.
A.
If both assertion and reason are true and the reason is the correct explanation of the assertion.
B.
If both assertion and reason are true but reason is not the correct explanation of the assertion.
C.
If assertion is true but reason is false.
D.
If the assertion and reason both are false.
E.
If assertion is false but reason is true.
Q180
Errorless
JEE Section
MCQ
Statement 1 : Atomic mass of sodium is $23u$ .
Statement 2 : An atom of sodium is 23 times heavier than an atom of carbon-12 isotope.
A.
If both assertion and reason are true and the reason is the correct explanation of the assertion.
B.
If both assertion and reason are true but reason is not the correct explanation of the assertion.
C.
If assertion is true but reason is false.
D.
If the assertion and reason both are false.
E.
If assertion is false but reason is true.
Q181
Errorless
JEE Section
MCQ
Statement 1 : Calomel is a chemical compound whereas brass is a mixture.
Statement 2 : Calomel always contains 5.6 times as much mercury as chlorine by weight. Brass can be made with widely different ratios of copper and zinc.
A.
If both assertion and reason are true and the reason is the correct explanation of the assertion.
B.
If both assertion and reason are true but reason is not the correct explanation of the assertion.
C.
If assertion is true but reason is false.
D.
If the assertion and reason both are false.
E.
If assertion is false but reason is true.
Q182
Errorless
JEE Section
MCQ
One litre of a mixture of $O_{2}$ and $O_{3}$ at NTP was allowed to react with an excess of acidified solution of KI. The iodine liberated required 40 mL of M/10 sodium thiosulphate solution for titration.
The volume of oxygen left unreacted in the mixture is
A.
0.655 L
B.
0.755 L
C.
0.855 L
D.
0.955 L
Q183
Errorless
JEE Section
MCQ
One litre of a mixture of $O_{2}$ and $O_{3}$ at NTP was allowed to react with an excess of acidified solution of KI. The iodine liberated required 40 mL of M/10 sodium thiosulphate solution for titration.
The weight percent of ozone in the mixture is
A.
5.57%
B.
6.57%
C.
7.57%
D.
8.57%
Q184
Errorless
JEE Section
MCQ
One litre of a mixture of $O_{2}$ and $O_{3}$ at NTP was allowed to react with an excess of acidified solution of KI. The iodine liberated required 40 mL of M/10 sodium thiosulphate solution for titration.
Knowing that ultraviolet radiations of wavelength 300nm can decompose ozone and assuming that one photon can decompose one molecule of ozone, the number of photons required for the complete decomposition of ozone in the original mixture will be
A.
$1.20 \times 10^{21}$
B.
$1.20 \times 10^{20}$
C.
$1.20 \times 10^{22}$
D.
$1.20 \times 10^{23}$
Q185
Errorless
JEE Section
MCQ
Nitric acid is the most important oxi-acid formed by nitrogen. It is one of the major industrial chemicals and is widely used. Nitric acid is manufactured by ostwald process in which catalytic oxidation of ammonia is done in following sequence as shown by reactions
$4 N H _ {3} (g) + 5 0 _ {2} (g) \xrightarrow [ \text {Catalyst} ]{\mathrm{Pt/Rh}} 4 N O (g) + 6 H _ {2} O (g)\tag{... (i}$
$2 N O (g) + O _ {2} (g) \xrightarrow {1 1 2 0 K} 2 N O _ {2} (g)\tag{... (ii}$
$3 N O _ {2} (g) + H _ {2} O (l) \longrightarrow 2 H N O _ {3} (a q) + N O (g) \quad \dots (\mathrm{iii})$
In this process the aqueous nitric acid is obtained which can be concentrated by distillation to $\sim$ 68.5% by weight. Then concentration to 98% acid can be achieved by dehydration with concentrated sulfuric acid.
85kg of $NH_{3}(g)$ was heated with 320kg oxygen in the first step and $HNO_{3}$ is prepared according to the above reactions. If the final solution has volume 500lt, then molarity of $HNO_{3}$ is
A.
3.33M
B.
8M
C.
2M
D.
6.66M
Q186
Errorless
JEE Section
MCQ
Nitric acid is the most important oxi-acid formed by nitrogen. It is one of the major industrial chemicals and is widely used. Nitric acid is manufactured by ostwald process in which catalytic oxidation of ammonia is done in following sequence as shown by reactions
$$
4 N H _ {3} (g) + 5 0 _ {2} (g) \xrightarrow [ \text {Catalyst} ]{\mathrm{Pt/Rh}} 4 N O (g) + 6 H _ {2} O (g)\tag{... (i}
$$
$$
2 N O (g) + O _ {2} (g) \xrightarrow {1 1 2 0 K} 2 N O _ {2} (g)\tag{... (ii}
$$
$$
3 N O _ {2} (g) + H _ {2} O (l) \longrightarrow 2 H N O _ {3} (a q) + N O (g) \quad \dots (\mathrm{iii})
$$
In this process the aqueous nitric acid is obtained which can be concentrated by distillation to $\sim$ 68.5% by weight. Then concentration to 98% acid can be achieved by dehydration with concentrated sulfuric acid.
If 180L of water completely reacts with $NO_{2}$ produced to form nitric acid according to the above reactions, then the volume of air required at STP containing 20% of $NH_{3}$ is
( $\rho_{H_{2}O} = 1g/mL$ )
A.
$1.56 \times 10^{6} L$
B.
$6.72 \times 10^{4} L$
C.
$3.36 \times 10^{6} L$
D.
None of these
Q187
Errorless
JEE Section
MCQ
Nitric acid is the most important oxi-acid formed by nitrogen. It is one of the major industrial chemicals and is widely used. Nitric acid is manufactured by ostwald process in which catalytic oxidation of ammonia is done in following sequence as shown by reactions
$$
4 N H _ {3} (g) + 5 0 _ {2} (g) \xrightarrow [ \text {Catalyst} ]{\mathrm{Pt/Rh}} 4 N O (g) + 6 H _ {2} O (g)\tag{... (i}
$$
$$
2 N O (g) + O _ {2} (g) \xrightarrow {1 1 2 0 K} 2 N O _ {2} (g)\tag{... (ii}
$$
$$
3 N O _ {2} (g) + H _ {2} O (l) \longrightarrow 2 H N O _ {3} (a q) + N O (g) \quad \dots (\mathrm{iii})
$$
In this process the aqueous nitric acid is obtained which can be concentrated by distillation to $\sim$ 68.5% by weight. Then concentration to 98% acid can be achieved by dehydration with concentrated sulfuric acid.
If 170kg of $NH_{3}$ is heated in excess of oxygen, then the volume of $H_{2}O(l)$ produced in 1st reaction at STP is
( $\rho_{H_{2}O}=1g/mL$ )
A.
$336\times10^{3}L$
B.
270L
C.
$224\times10^{3}L$
D.
170L
Q188
Errorless
Question Stem
Numerical
Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively). (Use Molar masses (in g mol $^{-1}$ ) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119 respectively).
The value of $x$ is
Correct Answer: 3.57
Explanation:
The value of $x$ is
Q189
Errorless
Matrix Match Type Questions
Numerical
Match the no. of moles listed in Column I(Number of moles) with their relevant amounts listed in Column II(Amount)
| Column-I | Column-II | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
|
Correct Answer: A→q,t;B→p;C→r;D→s
Explanation:
$\mathbf{A} \rightarrow \mathbf{q}, \mathbf{t}; \mathbf{B} \rightarrow \mathbf{p}; \mathbf{C} \rightarrow \mathbf{r}; \mathbf{D} \rightarrow \mathbf{s}$
A) $\because 2g = 1$ mole
∴ 2000mg = 1mole
$(\because 1g = 1000mg)$
∴ 200mg = $\frac{1}{2000}$ × 200 = 0.1mole
A) $\because$ mass of $1g Fe$ atom $= 56g = 1$ mole
So mass of $0.1g$ Fe atom $= \frac{56}{1} \times 0.1 = 5.6g$
$\because$ mass of $1g$ Fe atom $= 1$ mole (GAM of Fe)
∴ mass of 5.6g Fe atom = 0.1mole
B) $\because 22400ml$ of $CO_{2}$ at STP = 1mole of $CO_{2}$
∴ 4480ml of $CO_{2}$ at STP
$= \frac {1}{2 2 4 0 0} \times 4 4 8 0 = 0. 2 \text { mole of } C O _ {2}$
C) $\because 6.02 \times 10^{23}$ molecules of $O_2$ gas = 1 mole of $O_2$ gas
∴ $1.5 \times 10^{23}$ molecules of $O_{2}$ gas
$= \frac {1}{6 . 0 2 \times 1 0 ^ {2 3}} \times 1. 5 \times 1 0 ^ {2 3}$
= 0.25 mole of $O_{2}$ gas
D) $\because 18g$ of water $= 1$ mole
$\because$ density of water = 1g/ml
and mass of water = 18 g
$\begin{array}{r l} \therefore v & = \frac {m}{d} = \frac {1 8 g}{1 g} \times m l \\ & = 1 8 m l \end{array}$
Thus, 18ml of water = 1mole
$\begin{array}{r l} \therefore 9 m l \text { of water } & = \frac {1}{1 8 m l} \times 9 m l \\ & = 0.5 \text { mole }. \end{array}$
A) $\because 2g = 1$ mole
∴ 2000mg = 1mole
$(\because 1g = 1000mg)$
∴ 200mg = $\frac{1}{2000}$ × 200 = 0.1mole
A) $\because$ mass of $1g Fe$ atom $= 56g = 1$ mole
So mass of $0.1g$ Fe atom $= \frac{56}{1} \times 0.1 = 5.6g$
$\because$ mass of $1g$ Fe atom $= 1$ mole (GAM of Fe)
∴ mass of 5.6g Fe atom = 0.1mole
B) $\because 22400ml$ of $CO_{2}$ at STP = 1mole of $CO_{2}$
∴ 4480ml of $CO_{2}$ at STP
$= \frac {1}{2 2 4 0 0} \times 4 4 8 0 = 0. 2 \text { mole of } C O _ {2}$
C) $\because 6.02 \times 10^{23}$ molecules of $O_2$ gas = 1 mole of $O_2$ gas
∴ $1.5 \times 10^{23}$ molecules of $O_{2}$ gas
$= \frac {1}{6 . 0 2 \times 1 0 ^ {2 3}} \times 1. 5 \times 1 0 ^ {2 3}$
= 0.25 mole of $O_{2}$ gas
D) $\because 18g$ of water $= 1$ mole
$\because$ density of water = 1g/ml
and mass of water = 18 g
$\begin{array}{r l} \therefore v & = \frac {m}{d} = \frac {1 8 g}{1 g} \times m l \\ & = 1 8 m l \end{array}$
Thus, 18ml of water = 1mole
$\begin{array}{r l} \therefore 9 m l \text { of water } & = \frac {1}{1 8 m l} \times 9 m l \\ & = 0.5 \text { mole }. \end{array}$