Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

The amount of water obtained from combustion of $$16 grams$$ of methane is_______

A
16g

B
36g

C
32g

D
62g

Solution

$$CH_4(g)+2O_2\rightarrow CO_2(g)+2H_2O(g)$$
According to the above equation, one mole of $$CH_4(16\ g)$$ on combustion gives one mole of carbon dioxide and 2 moles of water and 2 moles of water $$=2\times(2+16)=36\ g$$ of water.
Question 2
1 / -0

Our body contains ___________ of water by weight.

A
$$70\%$$

B
$$60\%$$

C
$$50\%$$

D
$$80\%$$

Solution

Our body contains $$\underline{70\%}$$ of water by weights.
Question 3
1 / -0

The percentage weight by weight of $$0.5$$M $$CCl_4$$ solution in benzene is?
(Density of solution = $$1.4$$g/ml)

A
$$5.5\%$$

B
$$8.25\%$$

C
$$13\%$$

D
$$20.06\%$$
Question 4
1 / -0

If mass percentage of $$Mg^{2+}$$ in a biomolecule is 0.2%, then the number of $$Mg^{2+}$$ ions present in 10 g of biomolecule is (Atomic weight of Mg = 24)

A
$$\frac{1}{2} \times 10^{-2}N_{A}$$

B
$$\frac{1}{12} \times 10^{-2}N_{A}$$

C
$$\frac{1}{4}N_{A}$$

D
$$\frac{1}{12}N_{A}$$

Solution

$$100$$% of biomolecule contains-$$0.2$$% of $$Mg^{2+}$$
So, $$10g$$ of biomolecule contains-$$\cfrac {2}{100\times 10}\times 10$$
$$\Rightarrow2\times 10^{-2}gMg^{2+}$$
$$10g$$ of biomolecule contains-$$ 2\times 10^{-2}gMg^{2+}$$
$$\Rightarrow24g\longrightarrow N_A$$
$$\Rightarrow2\times 10^{-2}g\longrightarrow x$$
$$\Rightarrow x\longrightarrow \cfrac {2\times 10^{-2}g\times N_A}{24}$$
$$\Rightarrow x\longrightarrow \cfrac {1}{12}\times 10^{-2}N_A$$
The number of $$Mg^{2+}$$ ions present in $$10g$$ of biomolecules is $$\cfrac {1}{12}\times 10^{-2}N_A$$.
Question 5
1 / -0

$$H_2SO_4$$ is $$98\%$$ by weight of solution .Hence it is ?

A
1 molal

B
10 molal

C
50 molal

D
500 molal

Solution

Let the weight of solution be $$100g$$
Then the weight of $$H_2SO_4$$ will be $$98g$$
$$\therefore$$ Number of moles of $$98$$% $$H_2SO_4$$
$$=\cfrac {\text{Weight}}{\text{Molecular mass}}$$
$$=\cfrac {98}{98}=1$$
Molality= $$\cfrac {\text{No. of moles of solute}}{\text{Weight of solvent}}$$ (Weight of $$H_2O=100-98=2g$$)
$$=\cfrac {1}{2}$$
$$=\cfrac {1}{2}\times 1000=500$$ molal solution (Converting into kg)
Question 6
1 / -0

What is the proportion by weight the constitution elements of $$Cu$$ and $$O$$ in the compound $$CuO$$?

A
$$3.971: 1$$

B
$$4 : 2.968$$

C
$$4 : 3.971$$

D
$$2.968 : 2$$

Solution

Atomic mass of $$Cu=63.546\mu $$
Atomic mass of $$O=16\mu $$
In $$CuO$$, the proportion by weight is,
$$\dfrac { 63.546 }{ 16 } :\dfrac { 16 }{ 16 } \equiv 3.971:1$$
Question 7
1 / -0

Suppose 30% molecules have M = 20000; 40% molecules have M= 30000, rest of them have M=60000. then the PDI is :

A
1.45

B
1.20

C
0.83

D
0.98

Solution
Question 8
1 / -0

200 gram of a solution of water and sugar contains 20 g solute. What is the concentration of the solution in mass percentage?

A
20%

B
10%

C
40%

D
5%

Solution

Concentration of Solute = $$\dfrac{20}{200} \times 100$$ = 10 %
Question 9
1 / -0

How many grams of concentrated sulphuric acid $$ (80\%W/W) $$ sould be used to prepare 100 mL of one molar $$H_{2}SO_{4}$$ solution?

A
10 g

B
12. 25g

C
15 g

D
20 g

Solution
Question 10
1 / -0

$$40$$% of a mixture of $$0.2mol$$ of $${N}_{2}$$ and $$0.6mol$$ of $${H}_{2}$$ reacts to give $${NH}_{3}$$ according to the equation:
$${N}_{2}(g)+3{H}_{2}(g)\rightleftharpoons 2{NH}_{3}(g)$$
at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

A
$$4:5$$

B
$$5:4$$

C
$$7:10$$

D
$$8:5$$

Solution

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 56

Result

Some Basic Concepts of Chemistry Test - 56

Score

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SHARING IS CARING

Question 1

16g

36g

32g

62g

Solution

Question 2

$$70\%$$

$$60\%$$

$$50\%$$

$$80\%$$

Solution

Question 3

$$5.5\%$$

$$8.25\%$$

$$13\%$$

$$20.06\%$$

Question 4

$$\frac{1}{2} \times 10^{-2}N_{A}$$

$$\frac{1}{12} \times 10^{-2}N_{A}$$

$$\frac{1}{4}N_{A}$$

$$\frac{1}{12}N_{A}$$

Solution

Question 5

1 molal

10 molal

50 molal

500 molal

Solution

Question 6

$$3.971: 1$$

$$4 : 2.968$$

$$4 : 3.971$$

$$2.968 : 2$$

Solution

Question 7

1.45

1.20

0.83

0.98

Solution

Question 8

20%

10%

40%

5%

Solution

Question 9

10 g

12. 25g

15 g

20 g

Solution

Question 10

$$4:5$$

$$5:4$$

$$7:10$$

$$8:5$$

Solution

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