Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

$$800$$ g of a $$40\%$$ solution by weight was cooled. $$100$$ g of solute was precipitated. The percentage composition of remaining solution is :

A
$$31.4\%$$

B
$$20.0\%$$

C
$$23.0\%$$

D
$$24\%$$

Solution

$$\displaystyle Solute \: present \: in \: 800 \: g \: solution=800\times \frac{40}{100}$$$$=320 \: g$$

$$Solute \: precipitated = 100 \: g$$

$$Solute \: left =220 \: g$$

$$\therefore Total \: weight \: of \: solution = 800 - 100 = 700 \: g$$

$$700 \: g \: solution \: has \: solute =220$$

$$\displaystyle 100 \: g \: solution \: has \: solute=\frac{220}{700}\times
100= 31.43$$%
Question 2
1 / -0

In diammonium phosphate $$(NH_4)_2HPO_4$$, the percentage of $$P_2O_5$$ is :

A
35.87

B
46.44

C
51.99

D
53.78

Solution

Molar Mass of Diammonium Phosphate $$(NH_4)_2 HPO_4$$ is $$28+8+1+31+64=132 \ grams$$
Molar Mass of $$P_2O_5$$ is $$62+80=142 \ grams$$
We know that $$2$$ moles of diammonium hydrogen sulphate yields $$1$$ moles of $$P_2O_5$$ i.e $$2(NH_4)_2HPO_4 \rightarrow P_2O_5$$
Mass of $$2$$ moles of $$P_2O_5=2 \times 132 \ grams = 264 \ grams$$
Therefore, percentage composition of $$P_2O_5 = \cfrac {Mass \ of \ P_2O_5}{Mass \ of \ 2 \ moles \ of \ (NH_4)_2HPO_4}$$
$$=\cfrac {142}{264} \times 100 = 53.78$$%
Question 3
1 / -0

Atomic number (Z) of a neutral atom and mass number (A) of an atom are equal to:
(Here n = number of neutrons and p = number of protons):

A
Z = n and A = n+p

B
Z = e and A = n+e

C
Z = p and A = n+p

D
Z = n and A = p+e

Solution

The atomic number = Z = no. of protons = p
The mass no. = A = no. of protons (p) + no. of neutrons(n) = n + p
Question 4
1 / -0

How many g of $$KCl$$ would have to be dissolved in $$60$$ g $${H}_{2}O$$ to give $$20\%$$ by mass of solution?

A
$$15$$ g

B
$$1.5$$ g

C
$$11.5$$ g

D
$$31.5$$ g

Solution

As we know,
$$\%$$ by mass $$=\displaystyle\dfrac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$$

$$ 20 = \dfrac{\text{Mass of KCl} \times 100}{\text{Mass of KCl + Mass of water}}$$

Let, $$x$$ be the mass of KCl

$$20=\dfrac{100x}{x+60}$$

$$x=15 g$$

Therefore, mass of $$KCl$$ $$ = 15 $$ g

Hence, the correct option is $$A$$
Question 5
1 / -0

Calculate the volume occupied by 40 g. of a $$CH_4$$ at S.T.P if its V.D. is 8.

A
24 litres

B
46 litres

C
34 litres

D
56 litres

Solution

Molar mass $$=2\times vapour\ density = 2\times8 = 16 gm$$
No of moles $$=\frac{40}{16} = 2.5$$
So volume $$ = 2.5\times22.4 = 56\ litres$$
Question 6
1 / -0

The volume occupied by 7 grams of nitrogen gas at STP is:

A
6.5

B
5.6

C
7.5

D
4.5

Solution

Number of moles of nitrogen $$= \dfrac{7}{28} = \dfrac{1}{4} $$ moles

As we know, one mole occupies 22.4 litre volume.

Therefore, $$\dfrac{1}{4}$$ mole will occupy, $$V = \dfrac{22.4}{4} = 5.6\ litre$$
Question 7
1 / -0

Calculate the volume occupied by 0.01 moles of helium gas at STP.

A
2.24 L

B
0.224 L

C
22.4 L

D
None of the above

Solution

As we know, one mole of a gaseous substance occupies 22.4 litre, so volume occupied by 0.01 mole helium is $$0.224 \ L$$.
Question 8
1 / -0

Find the weight of a substance if its molecular weight is $$70$$ and in the gaseous form it occupies $$10$$ L at $$27^oC$$ and $$760$$ mm $$Hg$$ pressure.

A
32.5 g

B
35.5 g

C
28.38 g

D
None of the above

Solution

We know that 760 mm of hg is equal to 1 atm, so we use 1 atm.
As we know,
$$Pv=nRT\\ 1\times 10=\frac { w }{ 70 } \times 0.082\times 300\\ w=28.38g$$
Question 9
1 / -0

The percentage of $${P}_{2}{O}_{3}$$ in $${H}_{3}P{O}_{3}$$ is:

A
$$44.69$$

B
$$33.52$$

C
$$67.07$$

D
$$50.04$$

Solution

The reaction involved is as follows:
$${P}_{2}{O}_{3} + 3{H}_{2}O\rightarrow 2{H}_{3}P{O}_{3}$$
So, two moles of $$H_3PO_3$$ has one mole of $$P_2O_3$$.
$$\therefore \%$$ of $${P}_{2}{O}_{3}\ in\ {H}_{3}P{O}_{3} = \displaystyle\frac{110\times 100}{164} =$$ $$67.07\%$$
Question 10
1 / -0

$$4.4$$ g of $${N}_{2}O$$ occupies $$1.23$$ litres at $$2$$ atm and $$27^oC$$. The volume occupied by same mass of $$C{O}_{2}$$ at $$2$$ atm and $$27^oC$$ is :

A
2.24 L

B
2.46 L

C
1.12 L

D
1.23 L

Solution

Molecular mass of $$N_2O = CO_2 = 44\ g$$

Both have the same mass.

Therefore, no of moles are equal for both $$N_2O$$ and $$CO_2$$.

Avogadro's hypothesis states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles.

Hence, Volume of $$CO_2 =$$ Volume of $$N_2O$$

$$= 1.23$$ litres.

Hence, the correct answer is option $$\text{D}$$.

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Some Basic Concepts of Chemistry Test - 24

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Some Basic Concepts of Chemistry Test - 24

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SHARING IS CARING

Question 1

$$31.4\%$$

$$20.0\%$$

$$23.0\%$$

$$24\%$$

Solution

Question 2

35.87

46.44

51.99

53.78

Solution

Question 3

Z = n and A = n+p

Z = e and A = n+e

Z = p and A = n+p

Z = n and A = p+e

Solution

Question 4

$$15$$ g

$$1.5$$ g

$$11.5$$ g

$$31.5$$ g

Solution

Question 5

24 litres

46 litres

34 litres

56 litres

Solution

Question 6

6.5

5.6

7.5

4.5

Solution

Question 7

2.24 L

0.224 L

22.4 L

None of the above

Solution

Question 8

32.5 g

35.5 g

28.38 g

None of the above

Solution

Question 9

$$44.69$$

$$33.52$$

$$67.07$$

$$50.04$$

Solution

Question 10

2.24 L

2.46 L

1.12 L

1.23 L

Solution

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