A plant virus is found to consist of uniform symmetrical particles of 150 A o in diameter and 5000 A o long. The specific volume of the virus is 0 . 75 cm 3 g . If the virus is considered to be a single particle, if its molecular weight is x × 10 7 g then find x .

i) Volume of virus = πr 2 L (Volume of cylinder) (ii) Mass of single virus = Volume Specific Volume (iii) Molecular mass of virus = Mass of single virus × 6 . 023 × 10 23 Volume of virus = πr 2 L = 22 7 × 150 2 × 150 2 × 10 - 16 × 500 × 10 - 8 = 0 . 884 × 10 - 16 cm 3 Weight of one virus = 0 . 884 × 10 - 16 0 . 75 g = 1 . 178 × 10 - 19 g &there4; Mol. wt. of virus = 1 . 178 × 10 - 19 × 6 . 023 × 10 23 = 7 . 09 × 10 7 So the value of x = 7 . 09

Nitric acid can be produced from ammonia in three steps process. 4 NH 3 ( g ) + 5 O 2 ( g ) → 4 NO ( g ) + 6 H 2 O ( g ) . . . 1 2 NO g + O 2 g → 2 NO 2 g . . . 2 3 NO 2 g + H 2 O l → 2 HNO 3 aq . + NO g . . . 3 Calculate weight of NH 3 ( g ) required to produce 1260 kg of HNO 3 . When % Yield of 1 st , 2 nd and 3 rd reactions are respectively 69 % , 60 % and 68 % respectively.

As we can see that a series of three reactions are given in the question: 4 NH 3 ( g ) + 5 O 2 ( g ) → 4 NO ( g ) + 6 H 2 O ( g ) . . . 1 2 NO g + O 2 g → 2 NO 2 g . . . 2 3 NO 2 g + H 2 O l → 2 HNO 3 aq . + NO g . . . 3 The yield of every reaction is given in the question. The amount of ammonia required to produce 1260 kg of nitric acid can be calculated by stoichiometric calculation. The equation for the same is given below which is in accordance with the series of reactions. 1260 × 1000 63 = W NH 3 17 × 4 4 × 2 2 × 2 3 × 69 100 × 60 100 × 68 100 W NH 3 = 1811 . 59 kg

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Earn 100

When a mixture of $1$ -butene, $C_{4} H_{8}$ and butane $C_{4} H_{10}$ was burnt in excess of oxygen, $8.80 g$ of ${CO}_{2}$ and $4.14 g$ of $H_{2} O$ were obtained. What is the mass of butane in the mixture?

Important Questions on Some Basic Concepts of Chemistry

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>EXERCISE-3>Q 50

A plant virus is found to consist of uniform symmetrical particles of

150 \overset{o}{A}

in diameter and

5000 \overset{o}{A}

long. The specific volume of the virus is

\frac{0.75 {cm}^{3}}{g} .

If the virus is considered to be a single particle, if its molecular weight is

x \times 10^{7} g

then find

x

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>EXERCISE-3>Q 52

By the reaction of carbon and oxygen, a mixture of

CO

and

{CO}_{2}

is obtained. Find the difference in the mass of

CO

and

{CO}_{2}

in grams obtained when

20 grams

O_{2}

reacts with

12 grams

of carbon?

HARD

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>EXERCISE-3>Q 53

Nitric acid can be produced from ammonia in three steps process.

$4 {NH}_{3} (g) + 5 O_{2} (g) \to 4 NO (g) + 6 H_{2} O (g) . . . (1)$

$2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g) . . . (2)$

$3 {NO}_{2} (g) + H_{2} O (l) \to 2 {HNO}_{3} (aq .) + NO (g) . . . (3)$

Calculate weight of ${NH}_{3} (g)$ required to produce $1260 kg$ of ${HNO}_{3} .$ When $%$ Yield of $1^{st}, 2^{nd}$ and $3^{rd}$ reactions are respectively $69 %, 60 %$ and $68 %$ respectively.

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>EXERCISE-3>Q 54

A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the extent of

13.2 %

by weight.

2.58 g

of the mineral on heating lost

1.232 g

{CO}_{2} .

Calculate the

%

by weight of the other metal.

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>EXERCISE-3>Q 55

$6.2 g$ of a sample containing ${Na}_{2} {CO}_{3}, {NaHCO}_{3} &$ non-volatile inert impurity on gentle heating loses $5 %$ of its weight due to formation of $H_{2} O$ and ${CO}_{2}$ as in reaction

$2 {NaHCO}_{3} \to {Na}_{2} {CO}_{3} + H_{2} O + {CO}_{2}$ Residue is dissolved in water and formed $100 ml$ solution and its $10 ml$ portion requires $7.5 ml$ of $0.2 M$ aqueous solution of ${BaCl}_{2}$ for complete precipitation of carbonates. Determine weight (in gram) of ${Na}_{2} {CO}_{3}$ in the original sample

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Beta Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>EXERCISE-3>Q 56

An unknown metal reacts with excess chlorine to give the metal chloride

{MCl}_{2} .

When

2.375 gm

of metal chloride is dissolved in water and passed through an anion-exchange column packed with hydroxide ion, the solution requires

100 ml

0.5 M HCl

for neutralization. Calculate atomic mass (in

\frac{g}{mole}

) of unknown metal.

HARD

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>EXERCISE-3>Q 58

The density of mercury is

13.6 g / {cm}^{3} .

Estimate the

b

value.

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IMPORTANT

Beta Question Bank for Engineering: Chemistry>Chapter 1 - Some Basic Concepts of Chemistry>EXERCISE-3>Q 59

How many

mL

0.1 N HCl

are required to react completely with

1 g

mixture of

{Na}_{2} {CO}_{3}

and

{NaHCO}_{3}

containing equimolar amounts of two?

When a mixture of 1-butene, C4H8 and butane C4H10 was burnt in excess of oxygen, 8.80 g of CO2 and 4.14 g of H2O were obtained. What is the mass of butane in the mixture?

Important Questions on Some Basic Concepts of Chemistry

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When a mixture of $1$ -butene, $C_{4} H_{8}$ and butane $C_{4} H_{10}$ was burnt in excess of oxygen, $8.80 g$ of ${CO}_{2}$ and $4.14 g$ of $H_{2} O$ were obtained. What is the mass of butane in the mixture?