Chemistry Test 197

Result

Chemistry Test 197

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

Volume occupied by one molecule of water (density = 1 g/cm³) is :

A
5.5 × 10⁻²³ cm³

B
9 × 10⁻²³ cm³

C
6.023 × 10⁻²³ cm³

D
3 × 10⁻²³ cm³

Solution

Given: density of water is 1 g/cm³Gram Molecular mass of H₂O = 18 g/mole

Mass of one H₂O Molecule
Question 2
4 / -1

Liquid benzene (C₆H₆) burns in oxygen according to the equation,

2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(l)

How many litres of O₂ at STP are needed to complete the combustion of 39 g of liquid benzene? (Molecular weight of O₂ = 32, C₆H₆= 78)

A
74 L

B
11.2 L

C
22.4 L

D
84 L

Solution
Question 3
4 / -1

The solubility of a substance 'X' in pure ethanol is 0.1 gm/L and in water is 0.01 gm/L. To dissolve 11 gm of dry 'X'', 20 mL of fresh 50%5 (V/V) ethanol solution is added. How many times do we need to add this ethanol solution to dissolve 'X'?

A
100

B
10⁶

C
10³

D
10⁴

Solution
Question 4
4 / -1

In Haber's process, 30 litres of dihydrogen and 30 litres of dinitrogen were taken for the reaction, which yielded only 50% of the expected product. What will be the composition of the gaseous mixture under the aforesaid condition in the end?

A
20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen.

B
10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen.

C
20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen.

D
20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen.

Solution

(Limiting reagent is H₂ but only 50% conversion)

Final reaction mixture,

(10 L−NH₃, 25 L−N₂, 15 L−H₂)
Question 5
4 / -1

The molecular weight of glucose is 180. What is the molarity of a solution which contains 18 g/L of glucose?

A
0.1 M

B
0.2 M

C
0.3 M

D
0.4 M

Solution
Question 6
4 / -1

For the reaction represented by the equation, CX₄+ 2O₂→ CO₂+ 2X₂O, 9.0 g of CX₄ completely reacts with 1.74 g of oxygen. The approximate molar mass of X will be:

A
20

B
40

C
60

D
80

Solution

CX₄+ 2O₂ → CO₂ + 2X₂

According to the balanced reaction, one mole of the given compound must react with two moles of oxygen i.e., 64 g of oxygen.

∵ 1.74 g O₂ reacts with CX₄ = 9.0 g

∴ 64 g O₂will react with

This is the mass of one mole of the given compound i.e., molecular mass of CX₄ = 331 g

or 331 = 12 + 4X
Question 7
4 / -1

The number of atoms present in a 0.635 g of Cu piece will be

A
6.023 × 10⁻²³

B
6.023 × 10²³

C
6.023 × 10²²

D
6.023 × 10²¹

Solution

To calculate the number of atoms, we need to calculate the number of moles.

1 mole of atoms contain 6.023 × 10²³ atoms.

0.010.01 moles contain,

0.01N_A = 10⁻² × 6.023 × 10²³

= 6.023 × 10²¹ atoms.
Question 8
4 / -1

Which of the following has the smallest number of molecules?

A
0.1 mole of CO₂ gas.

B
11.2 L of CO₂ gas at STP.

C
22 g of CO₂ gas.

D
22.4 × 10³ mL of CO₂ gas at STP.

Solution

Equal numbers of moles have equal number of molecules.

Hence, the smallest number of molecules of CO₂ is in 0.1 mole of CO₂.
Question 9
4 / -1

In a solution of 7.8 g benzene (C₆H₆) and 46.0 g toluene (C₆H₅CH₃), the mole fraction of benzene is

A
1/2

B
1/3

C
1/5

D
1/6

Solution
Question 10
4 / -1

The amount of anhydrous Na₂CO₃ present in 250 mL of 0.25 M solution is

A
6.0 g

B
6.625 g

C
66.25 g

D
6.225 g

Solution