datavorous/entrance-exam-dataset huggingface dataset:
The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 20 mL of 0.01 M solution of CrH2O5ClCl2 as silver chloride is_______

Options:

The correct answer is:
4

Solution:
CrH2O5ClCl2⟶CrH2O5Cl+2Cl-

Number of ionisable Cl− in CrH2O5ClCl2â is 2.

Molarity of Cl– ions =0.01× 2=0.02 M

∴ Millimoles of Cl– ions in the given solution

=20×0.02

∴ Millimoles of AgNO3 required =20×0.2

V×0.1=20×0.02

V=4 mL

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The volume, in mL, of 0.02MK2Cr2O7, solution required to react with 0.288g of ferrous oxalate in acidic medium is………… (Molar mass of Fe=56gmol-1 )

Options:

The correct answer is:
100

Solution:
milliequivalent of K2Cr2O7= milliequivalent of FeC2O4

Vm1×0.02×6=0.288144×3×1000
Vm1×0.02×6=2×6
Vm1=100ml

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The volume (in ml) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [CrH2O)5Cl Cl2 , as silver chloride is close to

Options:
A 4
B 8
C 6
D 9

The correct answer is:
6

Solution:
Number of ionisable Cl- in [CrH2O)5ClCl2 is 2
Millimoles of Cl-=30×0.01×2=0.6
Millimoles of Ag+ required =0.6
0.6=0.1 V
V=6 ml

Topic: Chemistry, Coordination Compounds, JEE Main
The volume in $\mathrm{mL}$ of $0.1 \mathrm{M}$ solution of $\mathrm{NaOH}$ required to completely neutralise $100 \mathrm{~mL}$ of $0.3 \mathrm{M}$ solution of $\mathrm{H}_3 \mathrm{PO}_3$ is

Options:
A 60
B 600
C 300
D 30

The correct answer is:
600

Solution:
$\because$ Phosphorus acid $\left(\mathrm{H}_3 \mathrm{PO}_3\right)$ is a dibasic acid.
$\begin{aligned}
0.3 \mathrm{M} \mathrm{H}_3 \mathrm{PO}_3 & =0.6 \mathrm{~N} \mathrm{H}_3 \mathrm{PO}_3 \\
N_1 V_1 & =N_2 V_2 \\
0.1 \times V_1 & =0.6 \times 100 \\
V_1 & =600 \mathrm{~mL}
\end{aligned}
$

Topic: Chemistry, Redox Reactions, JEE Main
The volume occupied by 4.75 g of acetylene gas at 50°C and 740 mm Hg pressure is ____________ L. (Rounded off to the nearest integer)

[Given R=0.0826 L atm K-1 mol-1]

Options:

The correct answer is:
5

Solution:
Given mass=4.75 g⇒C2H2g

⇒Moles=4.7526 mol

Temperature=50+273=323 K

P=740760atm

R=0.0826ℓ atmmol K

⇒V=nRTP=4.7526×0.0826×323740760

⇒V=96314.07819240=5.0059ℓ≃5ℓ

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The volume of 0.02 M aqueous HBr required to neutralize 10.0 mL of 0.01 M aqueous Ba(OH)2 is (Assume complete neutralization)

Options:
A 2.5 mL
B 5.0 mL
C 10.0 mL
D 7.5 mL

The correct answer is:
10.0 mL

Solution:
The balanced chemical equation for the reaction between HBr and Ba(OH)2 is:

2HBr+Ba(OH)2 → BaBr2+2H2O

From this equation, we can see that 2 moles of HBr react with 1 mole of Ba(OH)2.

Equal equivalents will react.

Number of equivalents = Normality ×Volume

So,

N1V1=N2V20.02×V1=0.02×10V1=10 ml

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The volume of 0.1 M strong dibasic acid sufficient to neutralize 1g of a base that furnishes 0.04 mole of OH- in aqueous solution is :

Options:
A 400 mL
B 600 mL
C 200 mL
D 800 mL

The correct answer is:
200 mL

Solution:
For neutralisation;
⇒no. of moles of H+= no. of moles of OH-
⇒2×MaVa=0.04
⇒2×0.1×Va=0.04
⇒Va=0.040.2=0.2 L=200 mL

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2016 (10 Apr Online)
The volume of $10 \mathrm{~N}$ and $4 \mathrm{~N} \mathrm{HCl}$ required to make $1 \mathrm{~L}$ of $7 \mathrm{~N} \mathrm{HCl}$ are

Options:
A $0.50 \mathrm{~L}$ of $10 \mathrm{~N} \mathrm{HCl}$ and $0.50 \mathrm{~L}$ of $4 \mathrm{~N} \mathrm{HCl}$
B $0.60 \mathrm{~L}$ of $10 \mathrm{~N} \mathrm{HCl}$ and $0.40 \mathrm{~L}$ of $4 \mathrm{~N} \mathrm{HCl}$
C $0.80 \mathrm{~L}$ of $10 \mathrm{~N} \mathrm{HCl}$ and $0.20 \mathrm{~L}$ of $4 \mathrm{~N} \mathrm{HCl}$
D $0.75 \mathrm{~L}$ of $10 \mathrm{~N} \mathrm{HCl}$ and $0.25 \mathrm{~L}$ of $4 \mathrm{~N} \mathrm{HCl}$

The correct answer is:
$0.50 \mathrm{~L}$ of $10 \mathrm{~N} \mathrm{HCl}$ and $0.50 \mathrm{~L}$ of $4 \mathrm{~N} \mathrm{HCl}$

Solution:
Let $V$ litre of $10 \mathrm{~N} \mathrm{HCI}$ be mixed with $(1-V)$ litre of 4 $\mathrm{N} \mathrm{HCl}$ to give $(V+1-V)=1 \mathrm{~L}$ of $7 \mathrm{~N} \mathrm{HCl}$.
$$
\begin{aligned}
N_{1} V_{1}+N_{2} V_{2} &=N V \\
10 V+4(1-V) &=7 \times 1 \\
10 V+4-4 V &=7 \\
6 V &=7-4 \\
V &=\frac{3}{6}=0.50 \mathrm{~L}
\end{aligned}
$$
Volume of $10 \mathrm{~N} \mathrm{HCI}=0.50 \mathrm{~L}$
Volume of $4 \mathrm{~N} \mathrm{HCl}=1-0.50=0.50 \mathrm{~L}$

Topic: Chemistry, Solutions, JEE Main
The volume of a closed reaction vessel in which the following equilibrium reaction occurs is halved :
$2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})$
As a result,

Options:
A The rates of forward and backward reactions will remain the same.
B The equilibrium will not shift.
C The equilibrium will shift to the right.
D The rate of forward reaction will become double that of reverse reaction and the equilibrium will shift to the right.

The correct answer is:
The rate of forward reaction will become double that of reverse reaction and the equilibrium will shift to the right.

Solution:
$\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}$
If volume is reduced to half, $\mathrm{K}_{\mathrm{c}}$ will decrease to half. Thus, to maintain the equilibrium, the reaction should shift in the forward direction, i.e. towards right. Also, to attain equilibrium back, the rate of forward direction will become double the rate of backward direction.

Topic: Chemistry, Chemical Equilibrium, VITEEE, VITEEE 2019
The volume of a colloidal particle, $V_C$ as compared to the volume of a solute particle in a true solution $\mathrm{V}_{\mathrm{s}}$, could be

Options:
A
$\frac{V_{\mathrm{C}}}{\mathrm{V}_{\mathrm{S}}} \simeq 1$

B
$\frac{V_c}{V_{\mathrm{s}}} \simeq 10^{23}$

C
$\frac{V_{\mathrm{c}}}{\mathrm{V}_{\mathrm{S}}} \simeq 10^{-3}$

D
$\frac{V_c}{V_s} \simeq 10^3$

The correct answer is:
$\frac{V_c}{V_s} \simeq 10^3$

Solution:
No Solution Available

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2005
The volume of a cube is increasing at the rate of 9 cm3/sec. The rate (in cm2/sec) at which the surface area is increasing when the edge of the cube is 9 cm, is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Let, length of the edge of the cube is a, volume is V and surface area is S, then
V=a3⇒dVdt=3a2dadt
dadt=3a2 … 1
Now, S=6a2⇒dSdt=12adadt
dSdt=129×392=369=4cm2/sec.

Topic: Mathematics, Application of Derivatives, JEE Main
The volume of a cube is increasing at the rate of 18â cm 3 per second. When the edge of the cube is 12 cm, then the rate in cm2/s at which the surface area of the cube increases, is

Options:

The correct answer is:
6

Solution:
Let, x be the length of an edge, V be the volume and S be the surface area of the cube.
dVdt=18,V=x3âdVdt=18=3x2dxdtâdxdt=6x2â¯i
S=6x2
â dS dt =12x dx dt â¦ From i
â dS dt =12x 6 x 2
â dS dt =12Ã 6 x =6ââââ âµââGiven x=12
â´ dS dt =6â cm 2 /sec

Topic: Mathematics, Application of Derivatives, JEE Main
The volume of a gas at $20^{\circ} \mathrm{C}$ is $100 \mathrm{~cm}^3$ at normal pressure. If it is heated to $100^{\circ} \mathrm{C}$, its volume becomes 125 $\mathrm{cm}^3$ at the same pressure, then volume coefficient of the gas at normal pressure is

Options:
A $0.0015 /{ }^{\circ} \mathrm{C}$
B $0.0045 /{ }^{\circ} \mathrm{C}$
C $0.0025 /{ }^{\circ} \mathrm{C}$
D $0.0033 /{ }^{\circ} \mathrm{C}$

The correct answer is:
$0.0033 /{ }^{\circ} \mathrm{C}$

Solution:
$\frac{V_1}{V_2}=\frac{1+\gamma t_1}{1+\gamma t_2} \Rightarrow \frac{100}{125}=\frac{1+\gamma \times 20}{1+\gamma \times 100} \Rightarrow \gamma=0.0033 /{ }^{\circ} \mathrm{C}$

Topic: Physics, Thermal Properties of Matter, JEE Main
The volume of a metal sphere increases by $0.30 \%$, when its temperature is raised by $50^{\circ} \mathrm{C}$. The coefficient of linear expansion of the metal is

Options:
A $6 \times 10^{-5 /{ }^{\circ}} \mathrm{C}$
B $3 \times 10^{-5 /{ }^{\circ} \mathrm{C}}$
C $2 \times 10^{-5 /{ }^{\circ} \mathrm{C}}$
D $12 \times 10^{-5 /{ }^{\circ}} \mathrm{C}$

The correct answer is:
$2 \times 10^{-5 /{ }^{\circ} \mathrm{C}}$

Solution:
Coefficient of linear expansion $\alpha$ is about one third of the coefficient of volume expansion:
$\begin{aligned}
& \therefore \frac{\Delta V}{V}=(3 \alpha) \Delta T \\
& \Rightarrow 0.003=3 \alpha\left(50^{\circ} \mathrm{C}\right) \\
& \Rightarrow \alpha=2 \times 10^{-5} /{ }^{\circ} \mathrm{C}
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
The volume of a parallelopiped whose coterminous edges are $2 \overrightarrow{\mathbf{a}}, 2 \overrightarrow{\mathbf{b}}, 2 \overrightarrow{\mathbf{c}}$, is

Options:
A $2[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
B $4[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
C $8[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
D $[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$

The correct answer is:
$8[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$

Solution:
Volume of parallelopiped $=[2 \overrightarrow{\mathbf{a}} 2 \overrightarrow{\mathbf{b}} 2 \overrightarrow{\mathbf{c}}]$
$$
=8[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]
$$

Topic: Mathematics, Vector Algebra, JEE Main
The volume of a tetrahedron determined by the vectors aâ,bâ,câ is 34 cubic units. The volume (in cubic units) of a tetrahedron determined by the vectors 3aâÃbâ,4bâÃcâ and 5câÃaâ will be

Options:

The correct answer is:
202.5

Solution:
16a→ b→ c→=34⇒a→ b→ c→=92…1
Volume of tetrahedron =163a→×b→ 4b→×c→ 5c→×a→
=16×3×4×5a→ b→ c→2
=10×814=202.5 cubic units

Topic: Mathematics, Vector Algebra, JEE Main
The volume of air left in the lungs after a normal expiration is termed as:

Options:
A Residual volume
B Expiratory Reserve volume
C Expiratory capacity
D Functional Residual capacity

The correct answer is:
Functional Residual capacity

Solution:
Functional residual capacity (FRC = ERV + RV): It refers to the volume of air that remains in the lungs after a normal expiration.

Topic: Biology, Breathing and Exchange of Gases, NEET
The volume of air that remains in the lungs after a normal expiration is called ___.

Options:
A FRV
B FRC
C ERV
D RB

The correct answer is:
FRC

Solution:
The volume of air that remains in the lungs after a normal expiration is called FRC i.e. functional residual volume

Topic: Biology, Breathing and Exchange of Gases, NEET
The volume of an ideal gas $(\gamma=1.5)$ is changed adiabatically from 5 litres to 4 litres. The ratio of initial pressure to final pressure is:

Options:
A $\frac{16}{25}$
B $\frac{4}{5}$
C $\frac{8}{5 \sqrt{5}}$
D $\frac{2}{\sqrt{5}}$

The correct answer is:
$\frac{8}{5 \sqrt{5}}$

Solution:
For Adiabatic process
$\begin{aligned} & P_i V_i=P_f V_f^\gamma \\ & P_i(5)^{1.5}=P_f(4)^{1.5} \\ & \frac{P_i}{P_f}=\left(\frac{4}{5}\right)^{\frac{3}{2}}=\frac{4}{5} \cdot\left(\frac{4}{5}\right)^{\frac{1}{2}} \Rightarrow \frac{8}{5 \sqrt{5}}\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The volume of atoms (radius r) present in fcc unit cell of metal is :

Options:
A 203πr3
B 243πr3
C 123πr3
D 163πr3

The correct answer is:
163πr3

Solution:
Volume of atom in a cell =  4 3 πr 3  × N N = 4 for fcc
=43×πr3×4=163πr3

Topic: Chemistry, Solid State, JEE Main
The volume of $\mathrm{CO}_{2}\left(\mathrm{in} \mathrm{cm}^{3}\right)$ liberated at STP, when $1.06 \mathrm{~g}$ of anhydrous sodium carbonate is treated with excess of dilute $\mathrm{HCl}$ is [atomic mass of $\mathrm{Na}=23]$

Options:
A 112
B 224
C 56
D 2240

The correct answer is:
224

Solution:
$\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{HCl}$ (dil.) $\longrightarrow 2 \mathrm{NaCl}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}$
$1 \mathrm{~mol} \quad$ (dil.) $1 \mathrm{~mol}$
$=1.06 \mathrm{~g} \quad=22400 \mathrm{~cm}^{3}$
$1.06 \mathrm{~g} \mathrm{Na}_{2} \mathrm{CO}_{3}$ gives $=22400 \mathrm{~cm}^{3}$ of CO$_{2}$
$\therefore 1.06$ g gives $=\frac{1.06 \times 22400}{106}=224 \mathrm{~cm}^{3}$ of $\mathrm{CO}_{2}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than that of gas B at same temperature. What are the pressures of the gases for equal number of moles?

Options:
A 2PA=3PB
B PA=2PB
C 3PA=2PB
D PA=3PB

The correct answer is:
2PA=3PB

Solution:
PV=ZnRT
VA=2VB and ZA=3ZB
P=ZnRTV
PAPB=ZAVBZBVA=3ZB×VBZB×2VB=32
2PA=3PB
At constant T and moles; P∝ZV

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The volume of gas is reduced to half from its original volume. The specific heat will be

Options:
A
reduce to half

B
be doubled

C
remain constant

D
increase four times

The correct answer is:
remain constant

Solution:
Specific heat is an intensive property which depends only on the nature of the gas. Hence, if the volume of gas is reduced to half from its original volume the specific heat will remain constant.

Topic: Chemistry, Thermodynamics (C)
The volume of HCl, containing 73g L-1, required to completely neutralise NaOH obtained by reacting 0.69 g of metallic sodium with water, is _____ mL. (Nearest Integer)

(Given : molar Masses of Na,Cl,O,H are 23,35.5,16 and 1g mol-1 respectively)

Options:

The correct answer is:
15

Solution:
Mole of NaOH=0.6923=3×10-2

2Na+2H2O⟶2NaOH+H2

Number of moles of Na = Number of moles of NaOH

NaOH + HCl → NaCl + H2O

Moles of HCl = moles of NaOH

⇒ (Molarity × V)HCl = Number of moles of NaOH

⇒ 7336.5 × 1 × V = 0.03

⇒V = 15 × 10-3 Lit

⇒ V = 15 mL

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The volume of hydrogen liberated at STP by treating 2.4 g of magnesium with excess of hydrochloric acid is ......×10-2 L. Given Molar volume of gas is 22.4 L at STP. Molar mass of magnesium is 24 g mol-1

Options:

The correct answer is:
224

Solution:
The stoichiometric equation of the reaction between magnesium and hydrogen chloride can be written as follows,

One mole magnesium can liberate one mole of hydrogen gas according to the following equation.

Mg2.4 g=0.1 mole+2HCl→MgCl2+H20.1 mole
Volume of 1 mole H2 at =22.4 L
∴ 0.1 mole H2 at STP will occupy =2.24 L

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The volume of the greatest cone obtained by rotating a right-angled triangle of hypotenuse 2 units about a side is kÏ93 cubic units, then the value of k is equal to

Options:

The correct answer is:
16

Solution:
The volume generated =V=13πr2h
=13π4-x2x
For maximum volume, dVdx=0
⇒4-3x2=0⇒x=23

Vmax=π34-4323=16π93

Also, d2Vdx2<0

Topic: Mathematics, Application of Derivatives, JEE Main
The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius $=\sqrt{3}$ is:

Options:
A
$\frac{4}{3} \sqrt{3} \pi$

B
$\frac{8}{3} \sqrt{3} \pi$

C
$4 \pi$

D
$2 \pi$

The correct answer is:
$4 \pi$

Solution:
Given, radius of sphere $=\sqrt{3}$
Now, In $\triangle \mathrm{OAB}$, by Pythagoras theorem
$$
(\mathrm{OA})^2=(\mathrm{OB})^2+(\mathrm{AB})^2
$$
$$
\begin{aligned}
&(\sqrt{3})^2=\left(\frac{h}{2}\right)^2+r^2 \\
&3=\frac{h^2}{4}+r^2 \Rightarrow r^2=3-\frac{h^2}{4}
\end{aligned}
$$
Now, volume of cylinder $=\pi r^2 h$
$$
\begin{aligned}
&V=\pi\left(3-\frac{h^2}{4}\right) h \\
&V=3 \pi h-\frac{\pi h^3}{4}
\end{aligned}
$$
Now, for largest possible right circular cylinder the volume must be maximum
$\therefore$ For maximum volume, $\frac{d V}{d h}=0$ Now, Differentiating eq. (2) w.r.t. $h$ $V^1=\frac{d V}{d h}=3 \pi-\frac{3}{4} \pi h^2$
or $3 \pi-\frac{3}{4} \pi h^2=0 \Rightarrow 3 \pi=\frac{3}{4} \pi h^2$ $\Rightarrow h^2=4 \Rightarrow h=2$
Now, volume $(V)$ of the cylinder
$$
=\pi\left(3-\frac{h^2}{4}\right) h=\pi(6-2)=4 \pi
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2014 (11 Apr Online)
The volume of the parallelopiped whose edges are represented by $-12 \mathbf{i}+\alpha \mathbf{k}, 3 \mathbf{j}-\mathbf{k}$ and $2 \mathbf{i}+\mathbf{j}-15 \mathbf{k}$ is 546. Then $\alpha=$

Options:
A 3
B 2
C -3
D -2

The correct answer is:
-3

Solution:
Since $\left|\begin{array}{ccc}-12 & 0 & \alpha \\ 0 & 3 & -1 \\ 2 & 1 & -15\end{array}\right|=546 \Rightarrow \alpha=-3$.

Topic: Mathematics, Vector Algebra, JEE Main
The volume of the solid formed by rotating the area enclosed between the curve \( y=x^{2} \) and the line \( y=1 \) about \( y=1 \) is (in cubic units)

Options:
A \( \frac{9 \pi}{5} \)
B \( \frac{4 \pi}{3} \)
C \( \frac{8 \pi}{3} \)
D \( \frac{7 \pi}{5} \)

The correct answer is:
\( \frac{4 \pi}{3} \)

Solution:
Volume of the solid formed by rotating the area enclosed between the curve y=x2 and line y=1 will be V=∫012πxdy=2∫01πydy=4π3[y32]01=4π3 cu. units.

Topic: Mathematics, Area Under Curves, JEE Main
The volume of the tetrahedron having the edges \( \hat{i}+2 \widehat{j}-\hat{k}, \hat{i}+\widehat{j}+\hat{k}, \hat{i}-\widehat{j}+\lambda \widehat{k} \) as coterminous, is \( \frac{2}{3} \) cubic unit. Then \( \lambda \) equals

Options:
A \( 1 \)
B \( 2 \)
C \( 3 \)
D \( 4 \)

The correct answer is:
\( 1 \)

Solution:
Let a→=i⁡^+2j⁡^-k^, b→=i⁡^+j⁡^+k⁡^
and c→=i⁡^-j⁡^+λk⁡^
Since the volume of the tetrahedron =16a→ b→ c→
⇒ 23=1612-11111-1λ

Expand the determinant along R1

4=1λ+1-2(λ-1)-1(-1-1)
⇒ 4 = - λ + 5
⇒ λ = 1

Topic: Mathematics, Vector Algebra, JEE Main
The volume of the tetrahedron having the edges $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ as coterminous, is $\frac{2}{3}$ cubic unit. Then $\lambda$ equals

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
$\begin{aligned}
& \text { Let } \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& \text { and } \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}
\end{aligned}$
Since, volume of tetrahedron $=\frac{1}{6}[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$
$\begin{array}{ll}
\Rightarrow & \frac{2}{3}=\frac{1}{6}\left|\begin{array}{rrr}
1 & 2 & -1 \\
1 & 1 & 1 \\
1 & -1 & \lambda
\end{array}\right| \\
\Rightarrow & \frac{2}{3}=\frac{1}{6}[1(\lambda+1)-2(\lambda-1)-1(-1-1)] \\
\Rightarrow & 4=[-\lambda+5] \Rightarrow \lambda=1
\end{array}$

Topic: Mathematics, Vector Algebra, JEE Main
The volume of water to be added to \( 100 \mathrm{~cm}^{3} \) of \( 0.5 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4} \) to get decinormal concentration is

Options:
A \( 100 \mathrm{~cm}^{3} \)
B \( 450 \mathrm{~cm}^{3} \)
C \( 500 \mathrm{~cm}^{3} \)
D \( 400 \mathrm{~cm}^{3} \)

The correct answer is:
\( 400 \mathrm{~cm}^{3} \)

Solution:
The normality is inversely proportional to the volume of the solution.

Applying the equation,

N1V1=N2V2

given,

N1=0.5, N2=0.1V1=100 cm3, V2=?

0.5×100 =0.1×V2V2=500 cm3

Therefore, the water to be added to the 100 cm3 solution =500-100=400 cm3.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The volume percentage of Cl2 at equilibrium in the dissociation of PCl5 under a total pressure of 1.5atm is (Kp=0.202) ,

Options:
A 74.5
B 36.5
C 63.5
D 26.6

The correct answer is:
26.6

Solution:
PCl5âPCl3+Cl2
1-x x x
Total moles =1-x+x+x=1+x
k p = P PCl 3 â PCl 2 P PCl 5 = x 1+x P 1âx 1+x P x 1+x P
kP=x2P1-x2 1-x2=1
kP=x2P
x=kpP=0.2021.5=0.134=0.366
Mole â volume at constant P&T
Volume percentage =mole of Cl2totalmolesÃ100
=0.3661.366Ã100=26.86%

Topic: Chemistry, Chemical Equilibrium, JEE Main
The volume strength of 1.5 N H 2 O 2 is
(Given molar volume at STP = 22.4 L)

Options:

The correct answer is:
8.40

Solution:
Volume strength of H 2 O 2 =5.6×N
Volume strength =5.6×1.5
=8.4 volumes

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The volume strength of 1M H2O2 is:
(Molar mass of H2O2=34 g mol-1 )

Options:
A 5.6
B 22.4
C 11.2
D 16.8

The correct answer is:
11.2

Solution:
H2O2→H2O + 12O2
Volume strength=11.2×Molarity
=11.2×1
=11.2

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The volume strength of 8.9 MH2O2 solution calculated at 273 K and 1 atm is ........R=0.0821L atm K−1 mol−1 (rounded off to the nearest integer)

Options:

The correct answer is:
100

Solution:
Molarity of H2O2 solution =Volume strength11.2

Volume strength=8.9×11.2

=99.68 V

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The volume strength of 3.57 M solution hydrogen peroxide (at STP) is
Report your answer by rounding it upto nearest whole number

Options:

The correct answer is:
40

Solution:
Volume strength of H2O2 =11.2Ãmolarity

â volume strength =11.2Ã3.57â40

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The volume V of a given mass of monoatomic gas changes with temperature T according to the relation V=KT23. The workdone when temperature changes by 90 K will be xR. The value of x is
[R universal gas constant ]

Options:

The correct answer is:
60

Solution:
We know that work done is W=∫PdV ...1

⇒P=nRTV ...2

⇒W=∫nRTVdv ...3

and V=KT2/3 ...4

⇒W=∫nRTKT2/3·dv ...5

⇒ from 4 : dv=23KT-1/3dT

⇒W=∫T1T2nRTKT2/323 K1 T1/3dT

⇒W=23nR×T2-T1 ...6

⇒T2-T1=90 K ...7

⇒W=23nR×90

⇒W=60nR

Assuming 1 mole of gas

n=1

So W=60R

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (26 Feb Shift 2)
The volume V of an enclosure contains a mixture of three gases, 16 g of oxygen, 28 g of nitrogen and 44 g of carbon dioxide at absolute temperature T. Consider R as universal gas constant. The pressure of the mixture of gases is :

Options:
A 88RTV
B 3RTV
C 52RTV
D 4RTV

The correct answer is:
52RTV

Solution:
PV=n1+n2+n3RT

P×V=1632+2828+4444RT

PV=12+1+1RT

P=52RTV

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The volume-temperature graphs of a given mass of an ideal gas at constant pressures are shown below. What is the correct order of pressures?

Options:
A p1>p3>p2
B p1>p2>p3
C p2>p3>p1
D p2>p1>p3

The correct answer is:
p1>p2>p3

Solution:
Drawing a constant temperature line parallel to 'Y' axis, the isobar at higher volumes will be at lower pressure as P∝ 1 V at constant temperature.

Topic: Chemistry, States of Matter, JEE Main
The v-t graph of a moving object is given in the figure. The maximum acceleration is:

Options:
A $1 \mathrm{~cm} / \mathrm{sec}^2$
B $2 \mathrm{~cm} / \mathrm{sec}^2$
C $6 \mathrm{~cm} / \mathrm{sec}^2$
D $3 \mathrm{~cm} / \mathrm{sec}^2$

The correct answer is:
$6 \mathrm{~cm} / \mathrm{sec}^2$

Solution:
Acceleration in the 1st part,
$\begin{aligned} & a_1=\frac{20-0}{20}=1 \mathrm{~cm} / \mathrm{s}^2 \\ & a_2=0 \\ & a_3=\frac{80-20}{40-30}=\frac{60}{10}=6 \mathrm{~cm} / \mathrm{s}^2 \\ & a_4=\frac{80-0}{80-40}=\frac{80}{40}=2 \mathrm{~cm} / \mathrm{s}^2\end{aligned}$
Max. acceleration is $6 \mathrm{~cm} / \mathrm{s}^2$

Topic: Physics, Motion In One Dimension, JEE Main