Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

To convert molality into which of the following unit of concentration require density of the solution?

A
Percentage weight by weight

B
Percentage by volume

C
Mole fraction

D
Given all

Solution

Percentage of volume depends on the volume of the solution which also depends on the density. So the correct option is B
Question 2
1 / -0

25.5g of $$H_{2}O_{2}$$ solution on decomposition gave 1.68L of $$O_{2}$$ at STP. The percentage strength by weight of the solution is:

A
$$30$$

B
$$10$$

C
$$20$$

D
$$25$$

Solution

Balanced equation for decomposition of $$H_2O_2$$

$$2H_2O_2\rightarrow 2H_2O + O_2 $$

2mol $$H_2O_2$$ will produce 1 mol $$O_2$$

At STP 1 mol $$O_2$$ has volume = 22.4L

At STP 1.68L of $$O_2$$ = 1.68/22.4 = 0.075 mol $$O_2$$

Therefore the solution contained $$0.075\times2 = 0.15$$ mol $$H_2O_2$$

Molar mass $$H_2O_2 = 34g/mol$$

Mass of $$H_2O_2$$ decomposed $$= 0.15\times34 = 5.1$$ g $$H_2O_2$$ in 25.5g solution

100g solution contains : $$100/25.5\times5.1 = 20% $$

The original solution contains $$20% m/m H_2O_2$$

Hence, the correct option is $$C$$
Question 3
1 / -0

$$100$$gm of an aq. solution of sugar contains $$40\%$$ sugar by mass. How much water should be evaporated get $$50\%$$ sugar solution by mass?

A
$$10g$$

B
$$20g$$

C
$$0.0g$$

D
$$40g$$

Solution

$$\underline{\text{Initial condition}}:-$$ $$40\%$$ sugar (by mass),
Therefore, In $$100gm$$ of an aqueous solution, $$40g$$ sugar and $$60g$$ water is present.
$$\underline{\text{Final condition}}:-$$ $$50\%$$ sugar (by mass),
Therefore, In $$100gm$$ of an aqueous solution, $$50g$$ sugar and $$50g$$ water is present.
Amount of water to be evaporated $$=60-50=10\text{ }grams$$
Question 4
1 / -0

According to percentage weight arrange the following in descending order in the earth crust ?

A
$$O_2, Ca, Mg, S$$

B
$$O_2, S, Mg, Ca$$

C
$$S, Ca, Mg, O_2$$

D
$$Ca, O_2, Mg, S$$

Solution

Oxygen is present in maximum concentration in earth's crust while $$s$$ has less abundance,
So, abundance of availability of elements $$O_2>Ca>Mg>S.$$
Question 5
1 / -0

Calculate the mass percent of carbon in $${ C }_{ 2 }H_{ 5 }OH$$ (Molar mass of ethanol=46.068g)

A
$$13.13\%$$

B
$$88.79\%$$

C
$$52.14\%$$

D
$$34.73\%$$

Solution

Given : Mass of $$C_2H_5OH=46.08g$$
Molecular mass$$=46.08g$$
Percentage of carbon$$=\cfrac{\text{Mass of carbon}}{\text{Molecular mass}}\times 100\\=\cfrac{24}{46.08}\times 100\\=52.14\%$$
Question 6
1 / -0

One mole sample of $$FeO$$ is heated in air until it is completely converted into $$Fe_2O_3$$. The percentage increase in weight of the sample is?
[Given: Atomic mass of Fe = 56]

A
$$10.58\%$$

B
$$20.28\%$$

C
$$35.35\%$$

D
$$11.11\%$$

Solution

Mass of $$1$$ mol of $$FeO=(56+16)=72g$$

$$FeO+\cfrac{3}{4}O_2\longrightarrow \cfrac{1}{2}Fe_2O_3$$

$$1$$ mol of $$FeO$$ gives $$\cfrac{1}{2}$$ mol $$Fe_2O_3$$.

$$\therefore$$ Mass of sample after conversion $$=\cfrac{1}{2}\times M_{Fe_2O_3}=\cfrac{1}{2}\times 160=80g$$

$$\therefore \%$$ increase in weight $$=\cfrac{80-72}{72}\times 100=11.11\%$$

Hence, the correct option is $$\text{D}$$
Question 7
1 / -0

Aqueous urea solution is 20% by mass of solution. Calculate percentage by mass of solvent.

A
75%

B
15%

C
25%

D
65%

Solution

Let mass of the solution= $$100g$$
Given $$20$$% urea= $$20g$$ urea and $$80g$$ of water
$$\therefore$$ Mass of urea by solvent$$=\cfrac {20}{80}\times 100$$
$$=25$$%
Question 8
1 / -0

A gaseous mixture contains $$CH_4$$ and $$C_2H_6$$ in equimolecular proportion. The weight of 2.27 litres of this mixture at STP is:

A
3 g

B
4.6 g

C
1.6 g

D
2.3 g

Solution

Mol wt of $$CH_4=16\;g/mol$$

Mol wt of $$C_2H_6=30\;g/mol$$

Total $$=46\;g/mol$$

$$\Rightarrow 1$$ Mol wt of $$CH_4=22.4L$$

$$1$$ Mol wt of $$C_2H_6=22.4L$$

$$\Rightarrow 46\;g/mol$$ of $$CH_4+C_2H_6=22.4+22.4$$

$$=44.8L$$

$$\Rightarrow 46\;g/mol=44.8L$$

$$\Rightarrow x=2.27L$$

$$\Rightarrow x=\dfrac{46\times 2.27}{44.8}=2.33g$$

Hence, the answer is $$2.33g.$$

Hence, the correct option is $$D$$
Question 9
1 / -0

Which of the following is the best example of law of conservation of mass?
[n and m are the masses of reactants and p and q are the masses of products formed.]

A
$$n-m=p-q$$

B
$$n+m=p+q$$

C
$$n=m$$

D
$$p=q$$

Solution

Law of conservation of mass:- Mass is neither created nor destroyed it just from one compound to other.

$$n+m\rightarrow$$ $$p+q$$

Hence, the mass of reactants is equal to the mass of products formed.
Option $$B$$ is correct.
Question 10
1 / -0

$$27\ gm\ Al$$ is heated with $$49\ ml$$ of $$H_{2}SO_{4}$$ (specific gravity $$=2$$) produces $$H_{2}$$ gas. Percentage of $$Al$$ reacted with $$H_{2}SO_{4}$$ is:

A
$$50\%$$

B
$$66.66\%$$

C
$$75\%$$

D
$$80\%$$

Solution

$$2Al+3H_2SO_4\longrightarrow Al_2{(SO_4)}_3+3H_2\uparrow$$
$$2$$ moles of $$Al$$ reacts with $$3$$ moles of $$H_2SO_4$$.
i.e. $$2\times 27g$$ reacts with $$3\times 98g$$ $$H_2SO_4$$.
Mass of $$H_2SO_4$$ present $$=49\times 2=98g$$
$$\therefore $$ Mass of $$Al$$ reacted $$=\cfrac{2}{3}\times 27=18g$$
$$\therefore \%$$ $$Al$$ reacted $$=\cfrac{18}{27}\times 100=66.66\%$$

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 55

Result

Some Basic Concepts of Chemistry Test - 55

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SHARING IS CARING

Question 1

Percentage weight by weight

Percentage by volume

Mole fraction

Given all

Solution

Question 2

$$30$$

$$10$$

$$20$$

$$25$$

Solution

Question 3

$$10g$$

$$20g$$

$$0.0g$$

$$40g$$

Solution

Question 4

$$O_2, Ca, Mg, S$$

$$O_2, S, Mg, Ca$$

$$S, Ca, Mg, O_2$$

$$Ca, O_2, Mg, S$$

Solution

Question 5

$$13.13\%$$

$$88.79\%$$

$$52.14\%$$

$$34.73\%$$

Solution

Question 6

$$10.58\%$$

$$20.28\%$$

$$35.35\%$$

$$11.11\%$$

Solution

Question 7

75%

15%

25%

65%

Solution

Question 8

3 g

4.6 g

1.6 g

2.3 g

Solution

Question 9

$$n-m=p-q$$

$$n+m=p+q$$

$$n=m$$

$$p=q$$

Solution

Question 10

$$50\%$$

$$66.66\%$$

$$75\%$$

$$80\%$$

Solution

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