Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

A mixture of calcium and magnesium carbonates weighing 1.4 g was strongly heated until no further loss of weight was perceived. The residue weighed 0.76 g. What percentage of $$MgCO_3$$ was present in the mixture?

A
20.45

B
18.55

C
16.33

D
22.40

Solution

$$CaCO_3 + MgCO_3 \rightarrow CaO + MgO$$
let mass of $$CaCO_3$$ = x g
and mass of $$MgCO_3$$ = $$1.4-x$$ g

moles of $$CaCO_3$$ = $$\dfrac{x}{100}$$ = moles $$CaO$$ produced

moles of $$MgCO_3$$ = $$\dfrac{x}{84}$$ = moles $$MgO$$ produced

mass of $$CaO$$ produced = moles * no. of moles = $$\dfrac{(x) * 56}{100}$$

mass of $$MgO$$ produced = moles * no. of moles = $$\dfrac{(1.4-x) * 40}{100}$$

so, total weight of product = $$\dfrac{(1.4-x) * 56}{100}$$ + $$\dfrac{(1.4-x) * 40}{100}$$ = 0.76
so, value of x = 1.356 g
percentage of $$CaCO_3$$ = $$\dfrac{1.356}{1.4} * 100$$ = 96.8 %
percentage of $$MgCO_3$$ = 3.14 %
Question 2
1 / -0

The number of g-atom of oxygen in $$6.02\times { 10 }^{ 24 }$$ $$CO$$ molecule is

A
$$1$$

B
$$0.5$$

C
$$5$$

D
$$10$$

Solution

One molecule of $$CO$$ contains one oxygen atom
$$\therefore 6.02\times { 10 }^{ 24 }\quad $$ molecules of $$CO$$ contain $$6.02\times { 10 }^{ 24 }$$ oxygen atoms
$$6.02\times { 10 }^{ 23 }$$ atoms of oxygen $$\equiv$$ $$1g$$ atom of oxygen
$$\therefore 6.02\times { 10 }^{ 24 }$$ atoms of oxygen $$\equiv$$ $$10g$$ atom of oxygen.
Question 3
1 / -0

Law of constant composition is same as the law of:

A
conservation of mass

B
conversation of energy

C
multiple proportion

D
definite proportion

Solution

The law of constant composition is not the same as the law of conversation of mass or the law of conservation of energy, as these are related to the mass and energy of a system. It is also not related to the law of multiple proportions, which applies when two or more elements/ compounds have multiple ways of combining into different compounds.
The Law of Constant Composition, discovered by Joseph Proust, is also known as the Law of Definite Proportions.
According to the law of constant composition - "A chemical compound always contains its component element in the fixed ratio by mass." Hence option D is correct.
Question 4
1 / -0

Modern shoe soles are made of

A
Plastic

B
Synthetic rubber

C
Vulcanised rubber

D
Mixture of plastic and rubber
Question 5
1 / -0

If 100 gm of salt solution contains 20 g of salt dissolved in it, the percentage of the mass of the solution is________.

A
$$10\%$$

B
$$20\%$$

C
$$15\%$$

D
$$25\%$$

Solution

Given, $$100gm$$ solution contains $$20g$$ salt in it
$$\therefore$$ Percentage of mass in the solution= $$\cfrac {Mass\quad of\quad salt\quad dissolved}{Mass\quad of\quad solution}\times 100$$
$$=\cfrac {20gm}{100gm}\times 100$$
$$=20$$%
Question 6
1 / -0

How many grams of $$NaOH$$ will be required to prepare $$500\ g$$ solution containing $$10$$% $$\dfrac {w}{w}NaOH$$ solution?

A
$$100\ g$$

B
$$50\ g$$

C
$$0.5\ g$$

D
$$5.0\ g$$

Solution

In $$500\ g$$ of solution of 10% w/w $$NaOH$$ present.

Mass of the solute $$= \dfrac {10\times 500}{100} = 50\ g$$

So, $$50\ g$$ of $$NaOH$$ will be required to prepare $$500\ g$$ of $$10$$% $$\dfrac {w}{w}\ NaOH$$ solution.
Question 7
1 / -0

$$5.0\ g$$ sample of urea when heated with $$NaOBr$$ and $$NaOH$$ gives $$1120\ ml$$ of nitrogen at STP. The percentage purity of the sample is:

A
$$50$$%

B
$$40$$%

C
$$60$$%

D
$$30$$%

Solution

$${ NH }_{ 2 }CO{ NH }_{ 2 }+3NaOBr+2NaOH\longrightarrow { N }_{ 2 }+3NaBr+{ Na }_{ 2 }{ CO }_{ 3 }+3{ H }_{ 2 }O$$

$$\because$$ $$22400\ ml$$ of $${N}_{2}$$ is obtained from $$60\ g$$ of urea at S.T.P.

$$\therefore$$ $$1120\ ml$$ of $${N}_{2}$$ will be obtained from

$$\Rightarrow \cfrac { 60\times 1120 }{ 22400 } g$$ of urea $$=3\ g$$ or urea

Now,
$$\because$$ $$5\ g$$ sample contains $$3.0\ g$$ of urea

$$\therefore$$ $$100g$$ sample will contain $$\cfrac { 3.0\times 100 }{ 5.0 } =60g$$ of urea

Hence, the purity of urea sample is $$60$$%
Question 8
1 / -0

Sodium reacts with excess oxygen to form sodium oxide.A student wants to prepare 1.24 g of sodium oxide.While doing the calculations, he uses atomic number of sodium instead of obtained due to this mistake?

A
$$11\%$$

B
$$23\%$$

C
$$48\%$$

D
$$60\%$$
Question 9
1 / -0

The number of gram molecule of oxygen in $$6.02 \times 10^{24}$$ CO molecules is :

A
10 gm molecules

B
5 gm molecules

C
1 gm molecules

D
0.5 gm molecules

Solution

Given :

$$6.02\times 10^{24}$$ $$CO$$ molecules

We know,

$$1$$ mole $$CO$$ equivalent to $$6.023\times 10^{23}$$ molecules of $$CO$$

$$6.023\times 10^{23}$$ molecules of $$CO$$ contain $$6.023\times 10^{23}$$ atoms of $$O$$

Then, $$6.02\times 10^{24}$$ $$CO$$ molecules contain $$6.02\times 10^{24}$$ atoms of $$O$$

$$\Rightarrow$$ no. of gram atoms of $$O=\cfrac {6.02\times 10^{24}}{6.02\times 10^{23}}$$

$$=10gm$$ atoms $$\longrightarrow 1$$

As oxygen is a diatomic molecule.

No. of gm of molecules of oxygen=$$\cfrac {10gm\quad atoms}{2\quad atoms/molecule}=5gm$$ molecules

Therefore, the correct option is B.
Question 10
1 / -0

A solution has $$0.05\ M - Mg^{+2}$$ and $$0.05M - NH_{3}$$. Calculate the minimum concentration of $$NH_{4}Cl$$ required to prevent only precipitation of $$Mg$$ as $$Mg(OH)_{2}\cdot K_{sp}$$ of $$Mg(OH)_{2} = 9\times 10^{-12}$$ and $$K_{b}$$ of $$NH_{3} = 1.8\times 10^{-5}$$.

A
$$0.057\ M$$

B
$$0.067\ M$$

C
$$0.045\ M$$

D
$$1\ M$$

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 45

Result

Some Basic Concepts of Chemistry Test - 45

Score

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Rank

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SHARING IS CARING

Question 1

20.45

18.55

16.33

22.40

Solution

Question 2

$$1$$

$$0.5$$

$$5$$

$$10$$

Solution

Question 3

conservation of mass

conversation of energy

multiple proportion

definite proportion

Solution

Question 4

Plastic

Synthetic rubber

Vulcanised rubber

Mixture of plastic and rubber

Question 5

$$10\%$$

$$20\%$$

$$15\%$$

$$25\%$$

Solution

Question 6

$$100\ g$$

$$50\ g$$

$$0.5\ g$$

$$5.0\ g$$

Solution

Question 7

$$50$$%

$$40$$%

$$60$$%

$$30$$%

Solution

Question 8

$$11\%$$

$$23\%$$

$$48\%$$

$$60\%$$

Question 9

10 gm molecules

5 gm molecules

1 gm molecules

0.5 gm molecules

Solution

Question 10

$$0.057\ M$$

$$0.067\ M$$

$$0.045\ M$$

$$1\ M$$

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