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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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Density of dry air containing only $N_{2}$ and $O_{2}$ is 1.15 g/L at 740 mm and 300 K. What is % composition of $N_{2}$ by weight in the air?

78%

75.5%

70.02%

72.75%

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Explanation

$d = \frac{PM}{RT} 1.15 = \frac{(\frac{740}{760}) \times M}{0.0821 \times 300} ∴ M = 29.09 . Let mole fraction of N_{2} is x . 29.09 = 28 \times x + 32 (1 - x); x = 0.7275 wt . % = mole % \times \frac{mol . wt . of N_{2}}{average mol . wt . of air} = 72.75 \times \frac{28}{29.09} = 70.02$

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In preparation of iron from haematite $(F e_{2} O_{3})$ by the reaction with carbon
$F e_{2} O_{3} + C \to F e + C O_{2}$
How much 80% pure iron could be produced from 120 kg of 90% pure $F e_{2} O_{3}$ ?

94.5 kg

60.48 kg

116.66 kg

120 kg

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Explanation

Balanced reaction is

$2 F_{2} O_{3} + 3 C \to 4 F e + 3 C O_{2}$

No. of moles of $F e_{2} O_{3} = (\frac{120 \times 1000}{2 \times 56 + 48}) \times (\frac{90}{100})$

Mass of 80% pure iron produced

$= \frac{120 \times 1000 \times 0.9}{2 \times 56 + 48} \times \frac{2 \times 56}{0.8}$

= 94500 gram or 94.5 kg

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A gaseous mixture of $H_{2}$ and ${CO}_{2}$ gas contains 66 mass % of ${CO}_{2}$ . The vapour density of the mixture is:

6.1

5.4

2.7

10.8

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Explanation

$No . of moles of {CO}_{2} in 100 g mixture = \frac{66}{44} = 1.5 No . of moles of H_{2} in 100 g mixture = \frac{34}{2} = 17 M_{average} = \frac{1000}{18.5} = 5.40 V . D . = \frac{5.4}{2} = 2.7$

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The vapour density of a mixture containing ${NO}_{2}$ and $N_{2} O_{4}$ is 27.6. The mole fraction of $N_{2} O_{4}$ in the mixture is:

0.1

0.2

0.5

0.8

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Explanation

Let 1 mole of mixture has x mole $N_{2} O_{4}$

$2 \times 27.6 = x (92) + (1 - x) 46; x = 0.2$

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What volume of 75% alcohol by weight (d=0.80g/ $c m^{3}$ ) must be used to prepare 150 $c m^{3}$ of 30% alcohol by weight (d=0.90 g/ $c m^{3}$ )?

67.5 mL

56.25 mL

44.44 mL

None of these

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Explanation

Let V mL of alcohol be required

$∵$ mass of alcohol is same in both solutions

$∴$ $\frac{75}{100} \times 0.8 \times V = \frac{30}{100} \times 0.9 \times 150 V = 67.5 m L$

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Average atomic mass of magnesium is 24.31 a.m.u. This magnesium is compound of 79 mole% of ${}_{24}{Mg}$ and remaining 21 mole % of ${}_{25}{Mg}$ and ${}_{26}{Mg}$ . Calculate mole% of ${}_{26}{Mg}$ .

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Explanation

$Let mole % of {}^{26}{Mg} be x ∴ \frac{(21 - x) 25 + x (26) + 79 (24)}{100} = 24.31 x = 10 %$

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A mixture of $N H_{4} N O_{3} a n d (N H_{4})_{2} H P O_{4}$ contain 30.40% mass per cent of nitrogen. What is the mass ratio of the two components in the mixture?

2 : 1

1 : 2

3 : 4

4 : 1

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Explanation

$L e t w t . o f N H_{4} N O_{3} a n d (N H_{4})_{2} H P O_{4} a r e x a n d y g r a m r e s p e c t i v e l y \frac{\frac{x}{80} \times 2 \times 14 + \frac{y}{132} \times 2 \times 14}{x + y} \times 100 = 30.4 \Rightarrow x : y = 2 : 1$

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A mixture of $O_{2}$ and gas "Y" (mol. wt. 80) in the mole ratio a : b has a mean molecular weight 40. What would be mean molecular weight, if the gases are mixed in the ratio b : a under identical conditions? (gases are non-reacting):

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Explanation

$Let mole fraction of O_{2} is x 40 = 32 \times x + 80 (1 - x) or x = 5 / 6 a : b = x : (1 - x) = \frac{5}{6} : \frac{1}{6} When ratio is changed M_{Mixture} = 32 \times \frac{1}{6} + 80 \times \frac{5}{6} = 72$

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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.

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