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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

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Explanation

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

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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

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Explanation

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

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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

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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

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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

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Explanation

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

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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:

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Explanation

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

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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)

$4.98 \times 10^{- 21}$

$4.98 \times 10^{- 22}$

$6.55 \times 10^{- 21}$

$3.85 \times 10^{- 22}$

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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.

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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

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Explanation

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

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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%

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Explanation

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

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Rearrange the following (I to IV) in the order of increasing masses:

(I) 0.5 mole of O₃
(II) 0.5 gm atom of oxygen
(III) 3.011 $\times$ 10²³ molecules of O₂(IV) 5.6 litre of CO₂ at STP

II<IV<III<I

II<I<IV<III

IV<II<III<I

I<II<III<IV

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Explanation

$(I) 0.5 mole O_{3} = 24 g O_{3}; (II) 0.5 g atom of oxygen = 8 g (III) \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \times 32 = 16 g O_{2} (IV) \frac{5.6}{22.4} \times 44 g {CO}_{2} = 11 g {CO}_{2}$

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Nitric acid can be produced ${NH}_{3}$ in three steps process

(I) $4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (g)$
(II) $2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)$
(III) $3 N O_{2} (g) + H_{2} O (l) \to 2 H N O_{3} (a q) + N O (g)$

percent yield Ist, IInd and IIIrd are respectively 50%, 60% and 80% respectively then what volume of $N H_{3} (g)$ at 1 atm and 0 $°$ C required to produces 1575 g of $H N O_{3}$

156.25

350 L

3500 L

None of these

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Explanation

$M o l e s o f N O_{2} r e q u i r e d = (\frac{1575}{63}) \times \frac{3}{2} \times \frac{1}{0.8} = 46.875 m o l e s o f N O r e q u i r e d = \frac{46.875}{0.60} m o l e s o f N H_{3} r e q u i r e d = \frac{46.875}{0.60} \times \frac{1}{0.50} = 156.25 V o l u m e o f N H_{3} a t S T P r e q u i r e d = 156.25 \times 22.4 = 3500 L$

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Frequently Asked Questions

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

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

Chemistry MCQs for NEET — Practice Questions with Answers

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

Frequently Asked Questions

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