Blog Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

● ● ● ● ●

What volume of HCl solution of density 1.2 g/cm³ and containing 36.5% by weight HCl, must be allowed to react with zinc(Zn) in order to liberate 4.0 g of hydrogen?

333.33 mL

500 mL

614.66 mL

None of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$Zn + 2 HCl \overset{}{\to} {ZnCl}_{2} + H_{2}; moles of H_{2} evolved = 2 ∴ Moles of HCl required = 4 ∴ \frac{V \times 1.2 \times 0.365}{36.5} = 4; V = 333.33 mL$

● ● ● ● ●

What is the molar mass of diacidic organic Lewis base (B), if 12 g of chloroplatinate salt $({BH}_{2} {PtCl}_{6})$ on ignition produced 5 gm residue of Pt?

None of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$B H_{2} {PtCl}_{6} \overset{}{\to} Pt; \frac{12}{M_{B} + 410} = \frac{5}{195} = moles of Pt; Molecular mass of base = 58$

● ● ● ● ●

Equivalent weight of ${FeS}_{2}$ in the half reaction, ${FeS}_{2} \overset{}{\to} {Fe}_{2} O_{3} + {SO}_{2}$ is:

M/10

M/11

M/6

M/1

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$n = |1 + 5 \times 2|$

● ● ● ● ●

The equivalent weight of $HCl$ in the given reaction is: $K_{2} {Cr}_{2} O_{7} + 14 HCl \to 2 KCl + 2 {CrCl}_{3} + 3 {Cl}_{2} + H_{2} O$

16.25

36.5

85.1

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$14 mole {HCl}^{-} loses 6 mole e^{-}; ∴ 1 mole HCl loses \frac{6}{14} mole e^{-} ∴ eq . wt . of HCl = \frac{M}{(\frac{6}{14})} \Rightarrow \frac{36.5 \times 14}{6} = 85.1$

● ● ● ● ●

What volume of $O_{2} (g)$ measured at 1 atm and 273 K will be formed by action of 100 mL of 0.5 N KMnO₄ on hydrogen peroxide in an acid solution? The skeleton equation for the reaction is
${KMnO}_{4} + H_{2} {SO}_{4} + H_{2} O_{2}$ $\to K_{2} S O_{4} + M n S O_{4} + O_{2} + H_{2} O$

0.12 litre

0.028 litre

0.56 litre

1.12 litre

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$50 m ‐ eq . {KMnO}_{4} = 10 m ‐ mole {KMnO}_{4} or 25 m ‐ mole of O_{2} = 25 \times 22.4 \times 10^{- 3} = 0.56 L$

5H₂O₂+ 3H₂SO₄+ 2KMnO₄ = 5O₂+ 8H₂O + 2MnSO₄ + K₂SO₄

● ● ● ● ●

A solution of ${Na}_{2} S_{2} O_{3}$ is standardized iodometrically against 0.167 g of ${KBrO}_{3}$ . This process requires 50 mL of the ${Na}_{2} S_{2} O_{3}$ solution. What is the normality of the ${Na}_{2} S_{2} O_{3}$ ?

0.2 N

0.12 N

0.72 N

0.02 N

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$Eq . wt . of {KBrO}_{3} = \frac{1}{6} of its mol . wt . = \frac{1}{6} \times 167 N_{{Na}_{2} S_{2} O_{3}} = \frac{0.167}{167} \times 6 \times \frac{1}{0.05} = 0.12 N$

● ● ● ● ●

The ${NH}_{3}$ evolved due to complete conversion of N from 1.12 g sample of protein was absorbed in 45 mL of 0.4 N ${HNO}_{3}$ . The excess acid required 20 mL of 0.1 N $NaOH$ . The % N in the sample is:

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$milli ‐ equivalent of {NH}_{3} reacted with {HNO}_{3} = 45 \times 0.4 - 20 \times 0.1 = 16 ∴ \frac{W}{17} \times 1000 = 16; W_{{NH}_{3}} = 0.272 g; wt . of N = 0.272 \times \frac{14}{17} = 0.224 % N in the sample = \frac{0.224}{1.12} \times 100 = 20 %$

● ● ● ● ●

Cisplatin, an anticancer drug, has the molecular formula $Pt {({NH}_{3})}_{2} {Cl}_{2}$ . What is the mass (in gram) of one molecule? (Atomic weights : Pt=195, H=1.0, N=14, Cl=35.5)

$4.98 \times 10^{- 21}$

$4.98 \times 10^{- 22}$

$6.55 \times 10^{- 21}$

$3.85 \times 10^{- 22}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

The molecular weight of cisplatin is 300 g/mol (195 g/mol for Pt + 2 Ã— 14 g/mol for N + 6 Ã— 1 g/mol for H + 2 Ã— 35.5 g/mol for Cl). Converting to grams per molecule using Avogadro's number (6.022 Ã— 10^23 molecules/mol), we get 4.98 Ã— 10^-22 g/molecule.

● ● ● ● ●

The conversion of oxygen to ozone occurs to the extent of 15% only. The mass of ozone that can be prepared from 67.2 L oxygen at 1 atm and 273 K will be:

14.4 gm

96 gm

640 gm

64 gm

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Mole of $O_{2} = \frac{67.2}{22.4} = 3 m o l e 3 O_{2} \to 2 O_{3} ∴ M o l e o f O z o n e f o r m e d = \frac{2}{3} \times \frac{15}{100} \times 3 = 0.3 m o l e ∴ M a s s o f O z o n e f o r m e d = 0.3 \times 48 g = 14.4 g$

● ● ● ● ●

A silver coin weighing 11.34 g was dissolved in nitric acid. When sodium chloride was added to the solution all the silver (present as $A g N O_{3}$ ) was precipitated as silver chloride. The weight of the precipitated silver chloride was 14.35 g. Calculate the percentage of silver in the coin.

4.8%

95.2%

90%

80%

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$A g + H N O_{3} \to A g N O_{3} A g N O_{3} + N a C l \to N a N O_{3} + A g C l P O A C o n A g M o l e o f A g i n c o i n = m o l e o f A g i n A g C l a = \frac{14.35}{143.5} = 0.1 m o l e M a s s o f A g i n c o i n = 0.1 \times 108 = 10.8 g ∴ % s i l v e r i n c o i n = \frac{10.8}{11.34} \times 100 % = 95.2 %$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.

NEET Practice Questions (MCQs) with Answers & Solutions

5H2O2 + 3H2SO4 + 2KMnO4 = 5O2 + 8H2O + 2MnSO4 + K2SO4

Frequently Asked Questions

5H₂O₂+ 3H₂SO₄+ 2KMnO₄ = 5O₂+ 8H₂O + 2MnSO₄ + K₂SO₄