Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 42

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Question 1
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What could be better than a dozen $$(12)$$ donuts? How about a baker's dozen $$(13)$$ of donuts? Another large unit of measurement is known as Avogadro's number $$(6.022\times 10^{23})$$.
What is TRUE about Avogadro's number?

A
Avogadro's number is the number of particles in one mole of any element.

B
The molar mass of a substance will contain $$6.022\times 10^{23}$$ molecules.

C
There are $$6.022\times 10^{23}g$$ of carbon in $$12$$ mol of $$C - 12$$.

D
$$6.022\times 10^{23}$$ atoms of any element will have the exact same mass regardless of the identity of the element.

Solution

Avagadro constant i.e. $$6.023 \times 10^{23}$$ particles is the number of constituent particles, usually atoms or molecules, that are contained in the amount of substance given by $$1\space mole$$.
Question 2
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How many molecules of water are present in a $$0.25\ mole$$ of $$H_2O$$?

A
$$6.0\times{10}^{22}$$

B
$$4.5\times 10^{23}$$

C
$$1.5\times{10}^{23}$$

D
$$18\times 10^{23}$$

Solution

$$1\space mole$$ of a compound has $$6.023\times 10^{23}$$ molecules in it.
$$\Rightarrow 1\space mole =$$$$6.023\times 10^{23}$$ molecules
$$ 0.25\space mole = x$$ molecules
$$\Rightarrow x = (\dfrac{6.023}{4}) \times 10^{23}$$$$= 1.5\times 10^{23}$$ molecules of water.
Question 3
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What is the mass of $$6.022\times { 10 }^{ 23 }$$ molecules of $${NH}_{3}$$?

A
$$45.06g$$

B
$$19.06\ g$$

C
$$17.04g$$

D
$$31.02g$$

Solution

$$1\space mole$$ of $$NH_3$$ has $$6.023 \times 10^{23}$$ molecules of $$NH_3$$.
So, the mass of $$6.023 \times 10^{23}$$ molecules of $$NH_3$$. is the mass of $$1\space mole$$ of $$NH_3$$ i.e. molar mass of $$NH_3.$$
Molar mass $$= 14.01 + 3(1.01) = 17.04 \space g$$
Question 4
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What is the mass of $$6.022\times { 10 }^{ 23 }$$ formula units of $${({NH}_{4})}_{2}{SO}_{4}$$?

A
$$234.11\ g$$

B
$$132.11\ g$$

C
$$210.29\ g$$

D
$$342.14\ g$$

Solution

$$1\space mole$$ of $$(NH_4)_2SO_4$$ has $$6.023 \times 10^{23}$$ formula units.

So, the mass of $$6.023 \times 10^{23}$$ formula units of $$(NH_4)_2SO_4$$. is molar mass of $$(NH_4)_2SO_4$$

Molar mass of $$(NH_4)_2SO_4$$ $$= 2(14 + 4) + 32 + 4\times 16 = 132 \space g$$
Question 5
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Referring to the molecular sketch and table above, choose the option that best describes the percentage of carbon in the molecule, in terms of both atomic composition and mass-percent composition.

A
Carbon makes up approximately $$25$$% of the atoms, and $$25$$% of the mass.

B
Carbon makes up approximately $$40$$% of the atoms, and $$40$$% of the mass.

C
Carbon makes up approximately $$25$$% of the atoms, and $$40$$% of the mass.

D
Carbon makes up approximately $$40$$% of the atoms, and $$25$$% of the mass.

Solution

The molecular formula of compound $$= C_4H_9NO_3$$
$$\%$$ of carbon atom $$= \dfrac{4}{4+9+1+3} = \dfrac{4}{17} \times 100 \approx 25\%$$

$$\%$$ of carbon by mass $$= \dfrac{4\times 12}{48+9+14+48} = \dfrac{48}{119} \times 100 = 40.33\%$$
Question 6
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The atomic mass of copper is $$64\ amu$$, while the atomic mass of oxygen is $$16\ amu$$.
Let $$= copper$$ and $$= oxygen$$.
Which of the following would represent a compound that was $$50$$% copper, by mass?

A

B

C

D

Solution

Given, compound has $$50\%$$ copper, by mass.
Let Molar mass of compound $$= M$$
$$\Rightarrow $$ Mass of copper $$\dfrac{50}{100} \times M = 0.5 M$$
No. of moles of copper $$= \dfrac{0.5M}{64} ;$$
No. of moles of oxygen $$= \dfrac{0.5M}{16} ;$$
Ratio of copper to oxygen $$1:4$$
$$\Rightarrow CuO_4$$
Question 7
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Consider the figure below, in which the blue circles represent atoms of copper metal, atomic mass $$63.5\ amu$$, while the red circles represent atoms in zinc metal, atomic mass $$65.4\ amu$$.
Here are four statements concerning this system. Identify the correct statement or statements.
I. This is a single-phase mixture, with a copper-to-zinc atom ratio of $$3 : 1$$.
II. This is two-phase mixture, with a copper-to-zinc atom ratio of $$8 : 1$$.
III. The mixture is $$80$$% copper, by mass.
IV. This mixture could be an alloy of copper and zinc.

A
Statement I only.

B
Statement II only.

C
Statement I and III only.

D
Statement I and IV only.

Solution

Here, no. of atoms of $$Cu = 30\space atoms$$
No. of atoms of $$Zn = 10\space atoms$$
Single phase mixture is a homogenous mixture with same proportions of atoms.
Here, the ratio of copper to zinc is $$3:1$$
So, it is a two phase mixture.
$$\Rightarrow \% \space Cu $$ by mass $$= \dfrac{3\times 63.5}{3\times 63.5 + 65.4} = \dfrac{190.5}{255.9} = 0.744\times 100 = 74\% \approx 80\% $$ by mass
The mixture can be alloy of zinc.
Question 8
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How many molecules of methane are present in a sample that contains $$1.0\times { 10 }^{ 10 }$$ moles of methane?

A
$$6.0\times { 10 }^{ 23 }$$

B
$$1.0\times { 10 }^{ 33 }$$

C
$$6.0\times { 10 }^{ 33 }$$

D
$$6.0\times { 10 }^{ 24 }$$

Solution

$$1\space mole$$ of metane has $$6.023 \times 10^{23}$$ molecules of methane.

$$1\space mole = $$$$6.023 \times 10^{23}$$ molecules

$$10^{10}\space moles \rightarrow x$$

$$x = $$ $$(6.023 \times 10^{23})\times 10^{10} = $$ $$6.023 \times 10^{33}$$ molecules

So, $$10^{10}\space moles $$ has approximately $$6.0 \times 10^{33}$$ molecules
Question 9
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Equal masses of oxygen, hydrogen, and methane are taken in a container in identical condition. Find the ratio of volumes of the gases.

A
1: 32 :2

B
1 : 16: 2

C
1 : 1: 1

D
1: 16 :1

Solution

The molecular weights of a given mixture of gases are
1. Oxygen = 32
2. Hydrogen = 2
3. Methane = 16

Since masses are equal, assume mass as x,

So By using formula, no of moles = $$\dfrac{mass }{molecular weight}$$
$$n_{O_2} = \dfrac{x}{32}$$ : $$n_{H_2} = \dfrac{x}{2}$$ : $$n_{CH_4} = \dfrac{x}{16}$$

ratio is $$1 : 16 : 2$$
Question 10
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How many grams of $$CaWO_4$$ would contain the same mass of tungsten that is present in 569g of $$FeWO_4$ $? (W=184)

A
$$298$$ g

B
$$540$$ g

C
$$487$$ g

D
$$474$$ g

Solution

Let the mass of $$CaWO_4$$ be w g. As given,
Mass of W in w g of $$CaWO_4$$ = mass of W in 569g of $$FeWO_4$$

Moles of W in $$CaWO_4 \times $$at. mass of W = Moles of W in $$FeWO_4 \times $$ at. mass of W

As both $$CaWO_4$$ and $$FeWO_4$$ contains 1 atom of W each,
∴ moles of $$CaWO_4 \times $$ at. mass of W = Moles of $$FeWO_4 \times $$ at. mass of W
= $$\dfrac{w}{288} \times 184$$ = $$ (569/304) × 184$$
= $$W = 539.05 g.$$