Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

If 40 g of ethyl alcohol is dissolved in 50 ml of water, then calculate the weight/volume percentage of ethyl alcohol present in the solution? [Density of ethyl alcohol = 0.8 g/ml].

A
40%

B
30%

C
25%

D
35%

Solution

Weight/ Volume of component $$= \cfrac {Mass \ of \ component}{Total \ volume \ of \ solution} \times 100$$
Mass of component i.e ethyl alcohol $$=40 \ g$$
Volume of water $$=50 \ ml$$
Volume of ethyl alcohol $$= \cfrac {mass}{density}= \cfrac {40}{0.8}=50 \ ml$$
Total volume of solution $$= Volume \ of \ H_2O + Volume \ of \ ethyl \ alcohol$$
$$=50+50=100 \ ml$$
W/V of component $$= \cfrac {40}{100} \times 100 = 40$$%
Question 2
1 / -0

$$CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$$

The volume of CO$$_2$$ gas formed when 2.5 g calcium carbonate is dissolved in excess hydrochloric acid at 0$$^o$$C and 1 atm pressure is:
[1 mole of any gas at 0$$^o$$C and 1 atm pressure occupies 22.414 L volume].

A
1.12 L

B
56.0 L

C
0.28 L

D
0.56 L

Solution

$$CaCO_3 +2HCl \rightarrow CaCl_2 + H_2O + CO_2$$

100 g 44 g
1 mole 1 mole
22.4 L 22.4 L

We know that $$1$$ mole of any gas at $$0^oC$$ and $$1 \ atm$$ pressure occupies $$22.4 \ L$$ volume.

So, 100 g $$CaCO_3$$ forms $$22.4 \ L$$ of $$CO_2$$

Hence, 2.5 g $$CaCO_3$$ will form $$= \cfrac {2.5 \times 22.4}{100}$$

$$=\cfrac {56}{100} \ L$$ of $$CO_2$$

$$=0.56 \ L$$ of $$CO_2$$

Hence, the correct option is $$\text{D}$$
Question 3
1 / -0

A certain compound has the molecular formula, $$\displaystyle X_{4}O_{6}$$. If $$10$$ g of compound contains $$5.62$$ g of X, the atomic mass of X is approximately :

A
32 amu

B
30.8 amu

C
42 amu

D
48 amu

Solution

10 g of $$X_4O_6$$ contains $$5.62$$ g of X.

So, the weight of $$O$$ is $$10 - 5.62 = 4.38\:g$$.

Now, as $$\dfrac{weight}{atomic\: mass} =$$ number of atoms in given formula,

So, $$\dfrac{Weight\: of\: X}{atomic\: mass\: of\: X}$$ : $$\dfrac{Weight\: of\: O}{atomic\: mass\: of\: O}= 4:6$$

Let atomic mass of X be $$x$$.

$$\implies \dfrac{5.62}{x}$$ : $$\dfrac{4.38}{16} = 4:6$$

$$\implies \dfrac{5.62\times 16}{4.38x} = \dfrac{4}{6}$$

$$\implies \dfrac{5.62\times 16\times 6}{4} = 4.38x$$

So, $$x = 30.8 \: amu$$.
Question 4
1 / -0

The hydrated salt, $$\displaystyle Na_{2}SO_{4}.nH_{2}O$$ undergoes 56% loss in weight on heating and becomes anhydrous. The value of n (approx.) will be :

A
5

B
3

C
7

D
10

Solution

100 g of $$\displaystyle Na_{2}SO_{4}.nH_{2}O$$ contians 56 g of water and 44 g of $$\displaystyle Na_{2}SO_{4}$$.

The molar mass of $$\displaystyle Na_{2}SO_{4}$$ is $$23+23+32+4(16) = 142 \: g$$.

44 g of $$\displaystyle Na_{2}SO_{4}$$ corresponds to 56 g of water.

142 g of $$\displaystyle Na_{2}SO_{4}$$ will correspond to $$\displaystyle \frac {142 \times 56}{44}=180$$ g of water.

This corresponds to $$\displaystyle \frac {180}{18}=10$$ moles.

Thus, the value of n is 10.

Hence, option D is correct.
Question 5
1 / -0

What is the percentage of calcium in calcium carbonate $$(CaCO_3)$$?

A
30%

B
40%

C
50%

D
60%

Solution

Molecular weight of $${ CaCO }_{ 3 }=40+12+3\times 16$$
$$=52+48=100g$$
$$100gm$$ of $${ CaCO }_{ 3 }$$ contains $$40gm$$ of $$Ca$$
$$\therefore$$ % of $$Ca$$ in $${ CaCO }_{ 3 }$$ $$=\dfrac { 40 }{ 100 } \times 100=40$$%
Question 6
1 / -0

If a $$360 mL$$ sample of helium contains $$0.25$$ moles, how many molecules of chlorine gas would occupy the same volume at the same pressure and temperature?

A
$$1.2 \times {10}^{24}$$

B
$$6.022 \times {10}^{23}$$

C
$$3.01 \times {10}^{23}$$

D
$$1.51 \times {10}^{23}$$

E
$$7.55 \times {10}^{22}$$

Solution

For equal volumes of two gases at similar conditions of temperature and pressure, the number of moles of two gases will be same.

Hence, the number of moles of chlorine are 0.25.

When the number of moles of multiplied with Avogadro's number, the number of molecules are obtained.

Hence, the number of molecules of chlorine gas $$ = 0.25 $$ mol $$ \times 6.02 \times 10^{23} $$ molecules/mol $$ = 1.51\times 10^{23} $$ molecules.
Question 7
1 / -0

"All gases have the same number of moles in the same volume at constant temperature and pressure:. This statments belongs to :

A
Boyle's law

B
Charles's law

C
Avogadro's principle

D
ideal gas law

E
Dalton's law

Solution

Avogadro's law (sometimes referred to as Avogadro's hypothesis or Avogadro's principle) is an experimental gas law relating volume of a gas to the amount of substance of gas present.
Avogadro's law states that, "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".
For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.
Question 8
1 / -0

Statement 1: At the same temperature and pressure, 1 L of hydrogen gas and 1 L of neon gas have the same mass.
Statement 2: Equal volumes of ideal gases at the same temperature and pressure contain the same number of moles.

A
Both statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1

B
Both statement 1 and statement 2 are correct but Statement 2 is not the correct explanation of Statement 1

C
Statement 1 is correct but statement 2 is incorrect

D
Statement 1 is incorrect but statement 2 is correct

E
Both the statement 1 and statement 2 are incorrect

Solution

Statement 1: At the same temperature and pressure, 1 L of hydrogen gas and 1 L of neon gas have the same volume.
Statement 2: Equal volumes of ideal gases at the same temperature and pressure contain the same number of moles.
For example, 1 mole of any gas at STP will occupy a volume of 22.4 L.
Hence, statement 1 is not correct but statement 2 is correct.
Question 9
1 / -0

$$0.16\ g$$ of an organic compound containing sulphur produces $$0.233\ g$$ of $$BaSO_{4}$$. Percentage of sulphur in the compound is :

A
$$20$$

B
$$80$$

C
$$50$$

D
$$10$$

Solution

The molar masses of barium sulphate and sulphur are $$233\ g/mol$$ and $$32.1\ g/mol$$ respectively.

Number of moles of $$\displaystyle BaSO_4 = \dfrac {0.233}{233} = 0.001\ mol$$ $$\displaystyle = \text{number of moles of S.}$$

Mass of $$\displaystyle S = 0.001 \times 32.1 = 0.0321 \: g$$

Percentage of $$\displaystyle S = \dfrac {100 \times 0.0321}{0.16} = 20$$ %

Hence, option A is correct.
Question 10
1 / -0

How do you determine the mass percent of a solute in a solution?

A
$$ \displaystyle \text {Mass percent } = \dfrac { \text { mass of solute} }{ \text { mass of solution }} \times 100$$ %

B
$$ \displaystyle \text {Mass percent } = \dfrac { \text { mass of solution} }{ \text { mass of solute }} \times 100$$ %

C
$$ \displaystyle \text {Mass percent } = \dfrac { \text { moles of solute} }{ \text { mass of solution }} \times 100$$ %

D
$$ \displaystyle \text {Mass percent } = \dfrac { \text { mass of solute} }{ \text { moles of solution }} \times 100$$ %

E
$$ \displaystyle \text {Mass percent } = \dfrac { \text { moles of solute} }{ \text { moles of solution }} \times 100$$ %

Solution

$$ \displaystyle \text {Mass percent } = \dfrac { \text { mass of solute} }{ \text { mass of solution }} \times 100$$ %
Mass percentage is the grams of solute present in 100 gram of the solution.

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Some Basic Concepts of Chemistry Test - 69

Result

Some Basic Concepts of Chemistry Test - 69

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SHARING IS CARING

Question 1

40%

30%

25%

35%

Solution

Question 2

1.12 L

56.0 L

0.28 L

0.56 L

Solution

Question 3

32 amu

30.8 amu

42 amu

48 amu

Solution

Question 4

5

3

7

10

Solution

Question 5

30%

40%

50%

60%

Solution

Question 6

$$1.2 \times {10}^{24}$$

$$6.022 \times {10}^{23}$$

$$3.01 \times {10}^{23}$$

$$1.51 \times {10}^{23}$$

$$7.55 \times {10}^{22}$$

Solution

Question 7

Boyle's law

Charles's law

Avogadro's principle

ideal gas law

Dalton's law

Solution

Question 8

Both statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1

Both statement 1 and statement 2 are correct but Statement 2 is not the correct explanation of Statement 1

Statement 1 is correct but statement 2 is incorrect

Statement 1 is incorrect but statement 2 is correct

Both the statement 1 and statement 2 are incorrect

Solution

Question 9

$$20$$

$$80$$

$$50$$

$$10$$

Solution

Question 10

$$ \displaystyle \text {Mass percent } = \dfrac { \text { mass of solute} }{ \text { mass of solution }} \times 100$$ %

$$ \displaystyle \text {Mass percent } = \dfrac { \text { mass of solution} }{ \text { mass of solute }} \times 100$$ %

$$ \displaystyle \text {Mass percent } = \dfrac { \text { moles of solute} }{ \text { mass of solution }} \times 100$$ %

$$ \displaystyle \text {Mass percent } = \dfrac { \text { mass of solute} }{ \text { moles of solution }} \times 100$$ %

$$ \displaystyle \text {Mass percent } = \dfrac { \text { moles of solute} }{ \text { moles of solution }} \times 100$$ %

Solution

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