Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 27

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Weekly Quiz Competition

Question 1
1 / -0

If $$1$$ L of $$O_2$$ at $$15^oC$$ and $$750$$ mm pressure contains $$\text N$$ molecules, the number of molecules in $$2$$ L of $$SO_2$$ under the same conditions of temperature and pressure will be :

A
$$\dfrac {\text N}{2}$$

B
$$\text N$$

C
$$2\text N$$

D
$$4\text N$$

Solution

Under similar conditions of temperature and pressure, equal volume of gas contains equal number of molecules.
$$\therefore 1\ L=N$$ molecules
$$2\ L=2\ N$$ molecules
Question 2
1 / -0

Which of the following is the best thing to do during heavy lighting?

A
Lying on the ground in an open place

B
Going into the nearest water body

C
Staying indoors away from metallic doors or windows

D
Standing under a tall tree

Solution

During heavy lightning, the whole cloud fills up with electrical charges. A build up of positive charge builds up on the ground beneath the cloud, attracted to the negative charge in bottom of the cloud. The ground's positive charge concentrates around anything that sticks up - trees, windows, metallic doors and a spark of lightning strikes. So, one should stay away from metallic doors or windows as they are also charged.
Question 3
1 / -0

The percentage value of nitrogen in urea is about :

A
$$46$$

B
$$85$$

C
$$18$$

D
$$28$$

Solution

The formula of urea is $$\displaystyle NH_{2}CONH_{2}$$
$$\displaystyle W_{N}=28$$
$$\displaystyle W_{C}=12$$
$$\displaystyle W_{O}=16$$
$$\displaystyle W_{H}=4$$
In $$60$$ g urea, $$\displaystyle W_{N}=28g$$
$$100$$ g urea, $$\displaystyle W_{N}$$ is = $$\displaystyle \frac{28}{60}\times 100 = 46\%$$
Question 4
1 / -0

The weights of two elements which combines with one another are in the ratio of their:

A
atomic weight

B
molecular weight

C
equivalent weight

D
none of the above

Solution

According to the law of multiple proportions, when two elements combine to form more than one compound, the weight of one element combined with a fixed weight of the other is in a ratio of small whole numbers.
Hence, the weight of two elements that combine is in the ratio of their atomic weight.
So, the correct option is $$A$$.
Question 5
1 / -0

Modern atomic mass scale is based on the mass of:

A
$$H - 1$$

B
$$C - 12$$

C
$$C - 14$$

D
$$C - 16$$

Solution

Modern atomic weight scale is based on $$C - 12$$.
The standard unit for expressing the mass of atom is amu (atomic mass unit).
It is equal to 1/12 of the mass of an atom of carbon-12 equal to $$1.6605 \times 10^{-19} \: g$$.
amu is also called as avogram.
Avogram is a unit of mass and weight equal to one gram divided by the Avogadro's number.
Question 6
1 / -0

Insulin contains $$3.4$$% sulphur. The minimum molecular weight of insulin is :

A
941.176

B
944

C
945.27

D
none

Solution

$$3.4$$ g $$S$$ present in $$100$$ g insulin.
$$\therefore 32$$ g $$S$$ is present in $$\displaystyle \frac{100}{3.4}\times 32 = 941.176 $$ , which is equal to the minimum molecular weight of insulin.
Question 7
1 / -0

A sample of $$H_{2}SO_{4}$$ contains $$3.2$$ kg of sulphur. The weight (in g) of hydrogen present in the sample is:

A
100

B
200

C
50

D
150

Solution

In $$H_{2}SO_{4}$$ no of moles of hydrogen=2 $$\times$$ no of moles of hydrogen. Now 3.2kg of sulphur=3200 g of sulphur=100 moles of sulphur.
So, no fo moles of hydrogen=200 moles. So weight of the hydrogen=(1 $$\times$$ 200)=200 gm. So the correct option is B
Question 8
1 / -0

Percentage of Se in peroxidase anhydrous enzyme is $$0.5$$% by weight (At. wt. = $$78.4$$) then, minimum molecular weight of peroxidase anhydrous enzymes is :

A
$$\displaystyle 1.568\times 10^{4}$$

B
$$\displaystyle 1.568\times 10^{3}$$

C
$$15.68$$

D
$$\displaystyle 2.136\times 10^{4}$$

Solution

In peroxidase anhydrous enzyme 0.5% Se is present means, 0.5 g Se is present in 100 g of enzyme. In a molecule of enzyme one Se atom must be present. Hence 78.4 g Se will be present :

$$0.5$$ g Se present in $$100$$ g enzyme.

$$\therefore 78.4$$ g present in $$\displaystyle \frac{100}{0.5}\times 78.4 = $$Molecular weight
Question 9
1 / -0

A flask of methane $$(CH_4)$$ was weighed. Methane was then pushed out and the flask again weighed when filled with oxygen at the same temperature and pressure. The mass of oxygen would be :

A
the same as the methane

B
half of the methane

C
double of that of methane

D
negligible in comparison to that of methane

Solution

Avogadro's Law : It states "equal volumes of any two gases at the same temperature and pressure contain the same number of molecules". (same no of moles)

So both have same no of moles in flask,
$$n_1 = n_2$$

$$\displaystyle\frac{\omega_{CH_4}}{M_{CH_4}} = \displaystyle\frac{\omega_{O_2}}{M_{O_2}}$$

$$\displaystyle\frac{\omega_{O_2}}{\omega_{CH_4}} = \displaystyle\frac{M_{O_2}}{M_{CH_4}}= \displaystyle \frac {32}{16} = 2$$

So mass of oxygen would be double of that of methane.
Question 10
1 / -0

The density of a salt solution is $$1.13 g cm^{-3}$$ and it contains $$18\%$$ of $$NaCI$$ by weight The volume of the solution containing $$36.0\ g$$ of the salt will be

A
$$200 cm^3$$

B
$$217 cm^3$$

C
$$177 cm^3$$

D
$$157 cm^3$$

Solution

Consider the volume of the solution $$=x cm^3$$
Then the mass of the solution will be $$= 1.13x$$
$$(mass = density\times volume)$$
The solution contains 18% of NaCl by weight
$$\therefore \frac {18}{100}\times 1.13x=36$$
$$x=\frac {3600}{18\times 1.13}=177 cm^3$$

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 27

Result

Some Basic Concepts of Chemistry Test - 27

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SHARING IS CARING

Question 1

$$\dfrac {\text N}{2}$$

$$\text N$$

$$2\text N$$

$$4\text N$$

Solution

Question 2

Lying on the ground in an open place

Going into the nearest water body

Staying indoors away from metallic doors or windows

Standing under a tall tree

Solution

Question 3

$$46$$

$$85$$

$$18$$

$$28$$

Solution

Question 4

atomic weight

molecular weight

equivalent weight

none of the above

Solution

Question 5

$$H - 1$$

$$C - 12$$

$$C - 14$$

$$C - 16$$

Solution

Question 6

941.176

944

945.27

none

Solution

Question 7

100

200

50

150

Solution

Question 8

$$\displaystyle 1.568\times 10^{4}$$

$$\displaystyle 1.568\times 10^{3}$$

$$15.68$$

$$\displaystyle 2.136\times 10^{4}$$

Solution

Question 9

the same as the methane

half of the methane

double of that of methane

negligible in comparison to that of methane

Solution

Question 10

$$200 cm^3$$

$$217 cm^3$$

$$177 cm^3$$

$$157 cm^3$$

Solution

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