Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Calculate the percentage composition of nitrogen in urea ($$H_2NCONH_2$$).

A
46.6% $$N$$

B
43.6% $$N$$

C
44.6% $$N$$

D
42.6% $$N$$

Solution

Molecular weight of $${ H }_{ 2 }{ NCONH }_{ 2 }=4+28+12+16=60g$$
$$60g$$ of urea contains $$28g$$ of $$N$$
$$\therefore$$ % of $$N=\dfrac { 28 }{ 60 } \times 100=46.6$$%
Question 2
1 / -0

Calculate the percentage composition of iron in a ferric oxide ($$Fe_2O_3$$).
(Atomic mass of $$Fe = 56$$)

A
70% $$Fe$$

B
30% $$Fe$$

C
40% $$Fe$$

D
56.5% $$Fe$$

Solution

Molecular weight of $${ Fe }_{ 2 }{ O }_{ 3 }=56\times 2+16\times 3=112+48=160g$$
$$160g$$ of $${ Fe }_{ 2 }{ O }_{ 3 }$$ contains $$112g$$ of $$Fe$$
$$\therefore$$ % of $$Fe$$ $$=\dfrac { 112 }{ 160 } \times 100=70$$%
Question 3
1 / -0

What is the percentage composition of the elements in ammonia ($$NH_3$$)?
$$(N = 14,\ H = 1)$$

A
$$N= 82.35$$ %, $$H=17.65$$ %

B
$$N= 81.35$$ %, $$H=18.65$$ %

C
$$N= 80.35$$ %, $$H=19.65$$ %

D
$$N= 42.35$$ %, $$H=57.65$$ %

Solution

Molecular weight of $${ NH }_{ 3 }=17g$$
$$17g$$ of $${ NH }_{ 3 }$$ contains $$14g$$ of $$N$$ and $$3g$$ of $$H$$
$$\therefore$$ % of $$H=\dfrac { 3 }{ 17 } \times 100=17.65$$%
$$\therefore$$ % of $$N=\dfrac { 14 }{ 17 } \times 100=82.35$$%
Question 4
1 / -0

What is the mass of a mole of water containing 50% of heavy water $$(D_2O)$$ ?

A
$$19\ g$$

B
$$18\ g$$

C
$$20\ g$$

D
$$21\ g$$

Solution

A mole of water containing 100 % of $$ \displaystyle H_2O$$ weighs $$ \displaystyle 2(1)+16=18$$ g.
A mole of water containing 100 % of $$ \displaystyle D_2O$$ weighs $$ \displaystyle 2(2)+16=20$$ g.
A mole of water containing 50 % of $$ \displaystyle H_2O$$ and 50 % of $$ \displaystyle D_2O$$ weighs $$ \displaystyle \dfrac { 18+20}{2}=19$$ g.
Question 5
1 / -0

What are the percentage compositions of hydrogen and oxygen in water $$(H_2O)$$?
$$(H =1,\ O = 16)$$

A
11.1, 88.9

B
11.2, 88.8

C
11, 89

D
11.3, 88.7

Solution

Molecular weight of $${ H }_{ 2 }O=2+16=18g$$
$$18g$$ of $${ H }_{ 2 }O$$ contains $$2g$$ of $$H$$ and $$16g$$ of $$O$$
$$\therefore$$ % of $$H=\dfrac { 2 }{ 18 } \times 100=11.1$$%
% of $$O=\dfrac { 16 }{ 18 } \times 100=88.9$$%
Question 6
1 / -0

The value of the Avogadro constant is:

A
$$6.022 \times 10^{13}$$

B
$$6.022 \times 10^{22}$$

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

D
$$6.022 \times 10^{24}$$

Solution

Correct Answer: Option C
Explanation:

The value of the Avogadro constant is $$6.022 \times 10^{23}$$.

It is a number of atoms or molecules or ions or particles present in one mole of a substance.

Hence, the correct answer is option $$C$$.
Question 7
1 / -0

If 'M' is the molecular weight of a gas, what volume in $$l$$ at STP would be occupied by M/4 g of that gas?

A
11.2

B
2.24

C
5.6

D
22.4

Solution

If 'M' is the molecular weight of a gas, then $$\displaystyle M/4$$ g corresponds to $$\displaystyle 1/4$$ moles.

At STP, $$1$$ mole of gas occupies a volume of $$22.4$$ L.

At STP, $$\dfrac 14$$ mole of gas occupies a volume of $$\displaystyle \dfrac {22.4}{4}=5.6$$ L.

Hence, option C is correct.
Question 8
1 / -0

If 11g of oxalic acid are dissolved in 500 mL of solution (density= 1.1 g $$mL^{-1}$$), what is the mass % of oxalic acid in solution?

A
1%

B
2%

C
3%

D
4%

Solution

Mass of Oxalic acid (solute) = 11g
Mass of Solution = 500ml
Density = 1.1 g/ml
We know that
Mass % of solute = $$\dfrac{Mass of solute × 100}{Massofsolution}×50$$
Mass of solution = $$\text {Volume of solution} \times \text{Density of solution}$$
Mass of solution = $$500\times 1.1=50\times 11g $$
Mass % of the solute =$$\dfrac{11\times 100} {11\times 50}=2% $$
the answer is B
Question 9
1 / -0

The number of moles of NaCl in 3 litres of 3 M solution is:

A
1

B
3

C
9

D
27

Solution

molarity is equal to the number of moles of solute upon the volume of solution in liter.
$$MOLARITY = \dfrac{\text{number of moles }}{\text {volume of solution}}$$
number of moles = $$3\times 3$$ = 9
so the answer is 9 C option
Question 10
1 / -0

A solution is prepared by dissolving 5.64 g of glucose in 60 g of water. Calculate the mass percent of glucose.

A
8.59%

B
6.85%

C
9.34%

D
3.59%

Solution

mass percent of glucose = $$\dfrac{\text{mass of glucose}}{\text{mass of glucose +mass of water}} \times 100$$

= $$\dfrac{5.64}{5.64+60} \times 100$$
= 8.59 percent
answer is A

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 39

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Some Basic Concepts of Chemistry Test - 39

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Question 1

46.6% $$N$$

43.6% $$N$$

44.6% $$N$$

42.6% $$N$$

Solution

Question 2

70% $$Fe$$

30% $$Fe$$

40% $$Fe$$

56.5% $$Fe$$

Solution

Question 3

$$N= 82.35$$ %, $$H=17.65$$ %

$$N= 81.35$$ %, $$H=18.65$$ %

$$N= 80.35$$ %, $$H=19.65$$ %

$$N= 42.35$$ %, $$H=57.65$$ %

Solution

Question 4

$$19\ g$$

$$18\ g$$

$$20\ g$$

$$21\ g$$

Solution

Question 5

11.1, 88.9

11.2, 88.8

11, 89

11.3, 88.7

Solution

Question 6

$$6.022 \times 10^{13}$$

$$6.022 \times 10^{22}$$

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

$$6.022 \times 10^{24}$$

Solution

Question 7

11.2

2.24

5.6

22.4

Solution

Question 8

1%

2%

3%

4%

Solution

Question 9

1

3

9

27

Solution

Question 10

8.59%

6.85%

9.34%

3.59%

Solution

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