NEET Practice Questions

Q71.

When 7.3 g of magnesium bicarbonate (Mg(HCO₃)₂) is decomposed by heating, how much carbon dioxide gas (in mL) is released at STP?

1000 mL
1120 mL
2230 mL
3240 mL

Explanation:

Hint: The reaction is $Mg {({HCO}_{3})}_{2} \to {MgCO}_{3} + H_{2} O + {CO}_{2}$

Step 1:

The reaction is as follows:

Mg(HCO₃)₂ $\to$ MgCO₃ + H₂O +CO₂

No. of moles of Mg(HCO₃)₂ in 7.3 g

= \frac{7.3}{moles of 1 Mg {({HCO}_{3})}_{2}} = \frac{7.3}{146} = 0.05 moles

0.05 moles of Mg(HCO₃)₂ will generate 0.05 moles of CO₂

Step 2:

At STP, 1 mole of gas is 22.4 L or 22400 mL

Volume of

{CO}_{2} = \frac{5}{100} \times 22400 = 1120 mL

The decomposition temperature for MgCO₃ is 350 degree, and Here STP is written.

Q72.

In Young’s double-slit experiment, how many minimas can be obtained on a screen, if the wavelength of light used is 2000 $\overset{o}{A}$ and slit width is 0.0008 mm?

Explanation:

Path difference=dsin $θ = (2 n - 1) \frac{λ}{2}$

sin $θ$ $= (2 n - 1) \frac{λ}{2 d} = (2 n - 1) \frac{2000 \times 10^{- 10}}{2 \times 0.0008 \times 10^{- 3}} = \frac{(2 n - 1)}{8}$

$Maximum value of n = 4$
$So the total no . of minimas = 8 (4 above and 4 below the central maxima)$

Q73.

Which one of the following statements regarding post-fertilization development in flowering plants is incorrect?

Ovules develop into an embryo sac
Ovary develops into fruit
Zygote develops into an embryo
Central cell develops into endosperm

Q74.

The largest freezing point depression among the following 0.10 m solutions is shown by:

\$\\mathrm{KCl}\$
\$\\mathrm{C_6H_{12}O_6}\$
\$\\mathrm{Al}_2(\\mathrm{SO_4})_3\$
\$\\mathrm{K_2SO_4}\$

Explanation:

Hint: Higher the value of i, higher will be the depression in freezing point

KCl $\leftrightarrow K^{+}$ $+$ ${Cl}^{-}$

For KCl, i value is 2.

For $C_{6} H_{12} O_{6}$ , value of i is 1.

${Al}_{2} ({SO}_{4})_{3} \leftrightarrow 2 {Al}^{3 +}$ $+$ $3 {SO}_{4}^{2 -}$

For ${Al}_{2} ({SO}_{4})_{3}$ , i value is 5.

$K_{2} {SO}_{4} \leftrightarrow 2 K^{+}$ $+$ ${SO}_{4}^{2 -}$

For K₂SO₄, i vale is 3

For ${Al}_{2} ({SO}_{4})_{3}$ , i value is maximum hence, option third is the correct answer.

Q75.

In Duma’s method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is:

17.36
18.20
16.76
15.76

Explanation:

Hint: Weight of nitrogen = $ \dfrac{pVM} {RT}$

Step 1:
First, calculate the pressure of nitrogen as follows:
$ P_{N_{2}} \ = \ P_{Total} \ – \ P_{H_{2}O}$

= 725 – 25 = 700 mm

Now calculate the weight of nitrogen produced by using the PV =nRT formula as follows:

Weight of nitrogen = $\frac{pVM}{RT}$
where p is pressure= 700 mm
V is the volume of nitrogen = 40 mL
M is the molecular mass of N₂ = 28
R = 0.0821 Latm K^-1 mol^-1
T =300 K

Step 2:

On substituting,
Weight of nitrogen = $\frac{700 \times 40 \times 28}{760 \times 1000 \times 300 \times 0.0821}$
= 0.042 g

The amount of organic compound is 0.25. Calculate the percentage of nitrogen as follows:

% of nitrogen = $\frac{0.042}{0.25} \times 100 = 16.8$ $%$

Q76.

The energy liberated on complete fission of 1 kg of $U 92235$ is (Assume 200 MeV energy is liberated on fission of 1 nucleus)

(1) $8.2 \times 10^{10} J$

(2) $8.2 \times 10^{9} J$

(3) $8.2 \times 10^{13} J$

(4) $8.2 \times 10^{16} J$

Explanation:

(3) Mass of a uranium nucleus $= 92 \times 1.6725 \times 10^{- 27} + 143 \times 1.6747 \times 10^{- 27} = 393.35 \times 10^{- 27} kg$

Number of nuclei in the given mass $= \frac{1}{393.35 \times 10^{- 27}} = 2.542 \times 10^{24}$

Energy released $= 200 \times 2.542 \times 10^{24} MeV$

$= 5.08 \times 10^{26} MeV = 8.135 \times 10^{13} J = 8.2 \times 10^{13} J$

Q77.

A certain mass of hydrogen is changed to helium by the process of fusion. The mass defect in the fusion reaction is 0.02866 u. The energy liberated per nucleon is: (given 1 u = 931 MeV)

67 MeV
26.7 MeV
6.675 MeV
13.35 MeV

Q78.

A beam of light of 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is

Explanation:

$F r o m t h e f i g u r e, \tan θ = \frac{\frac{x}{2}}{2} = \frac{x}{4}$
$F o r s m a l l θ a n d w h e n θ i s c o u n t e d i n r a d, \tan θ \approx θ \Rightarrow θ = \frac{x}{4}$
$\Rightarrow \frac{λ}{a} = \frac{x}{4}$
$\Rightarrow x = \frac{4 λ}{a} = 2.4 m m$

Q79.

What is the work done when one mole of an ideal gas undergoes isothermal expansion from an initial pressure of 10 atm to a final pressure of 1 atm at a constant temperature of 300 K? (Gas constant = 2 calorie )

938.8 cal
1138.8 cal
1381.8 cal
1581.8 cal

Explanation:

(3) $- W = + 2 .303 n R T$ log $\frac{p_{1}}{p_{2}}$

$- W = 2 .303 \times 1 \times 2 \times 300$ log $\frac{10}{1} = 1381 .8 c a l .$

NCERT Reference: Page Number: 160, Paragraph: 6.2.1

Q80. What causes the first heart sound (lub) during a heartbeat?

Closure of the semilunar valves at the start of ventricular diastole
Opening of the semilunar valves to allow blood flow to the arteries
Rapid filling of the ventricles from the atria
Closure of the tricuspid and bicuspid valves during ventricular systole

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