Atoms and Molecules Test

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Atoms and Molecules Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

Insulin contains 3.4% sulphur by mass. What will be minimum molecular weight of insulin?

A
$$94.117$$

B
$$1884$$

C
$$941.7$$

D
$$976$$

Solution

The formula for mass percentage is given by:
$$Mass\ percentage = \dfrac {mass\ of\ solute}{mass\ of\ solution}$$

We need to find the molar mass of solution (insulin). We will take the molar mass of solute (sulphur) $$= 32\ g/mol$$

Mass percentage is given as $$3.4\%$$

Substituting these values in the formula gives:

$$\dfrac {3.4}{100} = \dfrac {32}{mass\ of\ solution}$$

Mass of solution (insulin) $$= \dfrac {32}{3.4}\times 100 = 941.7\ g/mol$$
Question 2
1 / -0

The number of molecule in $$4.25$$g of $$NH_3$$ is:

A
$$1.505\times 10^{23}$$

B
$$3.01\times 10^{23}$$

C
$$6.02\times 10^{23}$$

D
none of these

Solution

Molar mass $$NH_3$$ = 14+3 = 17g/mol
Given mass = 4.25g
No of moles= $$\dfrac{given\ mass}{molar\ mass}$$
No of moles = 4.25/17 = 0.25 mol of  $$NH_3$$
1 mol of  $$NH_3$$ contains  $$6.022\times 10^{23}$$ molecules
0.25 mol $$NH_3$$ contains $$0.25 \times$$$$6.022\times 10^{23}$$  = $$1.5055\times$$ $$10^{23}$$ molecules
Question 3
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A compound contains $$10^{-2}\%$$ of phosphorus. If the atomic mass of P is $$31$$, then the molecular mass of the compound having one phosphorous atom per molecule is:

A
$$3.1\times 10^5$$

B
$$3.1$$

C
$$31$$

D
None of these
Question 4
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The Avogadro number ($$N_A$$), is changed from $$6.022\times 10^{23}$$ $$mol^{-1}$$ to $$6.022\times 10^{20}$$ $$mol^{-1}$$, this would change:

A
the ratio of chemical species to each other in a balanced equation

B
the ratio of elements to each other in a compound

C
the definition of mass in units of grams

D
the mass of one mole of carbon

Solution

Let us consider the Avogadro number ($$N_A$$) is changed from $$6.022 \times {10^{23}}\,mo{l^{ - 1}}\,\ to \ 6.022 \times {10^{20}}\,\,mo{l^{ - 1}}$$ this would be change the mass of one mole of carbon.
Hence,
mass of one mole of carbon $$=12$$ gram
or it is a mass of $$6.022 \times {10^{23}}$$ atoms of carbon.
Therefore,
Now, the mass of $$6.022 \times {10^{20}}$$ atoms of carbon

$$=\dfrac{{12}}{{6.022 \times {{10}^{23}}}} \times 6.022 \times {10^{20}}$$

$$=0.012\,\,g$$

So, the correct option is D.
Question 5
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A sample of potato starch was ground in a ball mill to give a starch-like molecule of lower molecular weight. The product analysed was 0.086 % phosphorus. If each molecule is assumed to contain one atom of phosphorus, what is the molecular mass of the material?

A
$$3.42\times 10^4$$ amu

B
$$3.72\times 10^4$$ amu

C
$$3.6\times 10^4$$ amu

D
None of these

Solution

Since the percentage of $$P$$ is $$0.086$$

$$\therefore$$ $$0.086g$$ of $$P$$ will be present in $$100g$$ of starch.

Then, $$1 g $$ of $$P$$ will be present in $$\dfrac{100}{0.0866}$$ g of starch

The atomic mass of $$P$$ is $$31 \ a.m.u$$.

$$31 \ a.m.u$$ of $$P$$ will be contained in $$\dfrac{100\times 31}{0.0866} \ a.m.u \ $$ of Starch.

$$\therefore$$ Molecular weight of material= $$3.6 \times 10^4 \ a.m.u$$
Question 6
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A compound contains $$4$$% oxygen, then the minimum molecular weight of that compound will be:

A
$$400g$$

B
$$800g$$

C
$$200g$$

D
$$100g$$

Solution

So $$1$$ mole of oxygen $$=16\ grams$$
For the Molecular weight of the molecule to be least, no of oxygen moles should be $$1$$.

Let's assume that there are $$16$$ gms of oxygen present in the compound, which should be $$4$$% of the total molecular mass.

$$\dfrac { 4x }{ 100 } =16\ gm$$

$$ 4x =1600\ gm$$

Therefore,$$x=400 \ grams$$
Hence, option $$A$$ is correct.
Question 7
1 / -0

The number of atoms of the $$He$$ in $$104\ amu$$ is:

A
$$3.1 \times 10^{25}$$

B
$$6.2 \times 10^{25}$$

C
$$26$$

D
$$206$$

Solution

Since the mass of one atom of He = 4 a.m.u
Therefore the number of atoms of the He in 104 amu is, $$\dfrac { 104 }{ 4 } =26$$.
Question 8
1 / -0

The number of atoms of oxygen present in 10.6 g of $$Na_2CO_3$$ will be:

A
$$6.02 \times 10^{22}$$

B
$$12.04 \times 10^{22}$$

C
$$1.806 \times 10^{23}$$

D
$$31.8$$

Solution

Molar mass of $$Na_2CO_3$$ $$= (2 \times atomic\ mass\ of\ nitrogen) + (1 \times atomic\ mass\ of\ carbon) + (3 \times atomic\ mass\ of\ oxygen)$$ $$= 2 (23) + 1 (12) + 3 (16)$$ $$= 106\ g/mol$$

So, $$1$$ mole of $${ Na }_{ 2 }{ CO }_{ 3 }=108g$$

Moles in $$10.6g$$ of $${ Na }_{ 2 }{ CO }_{ 3 }=10.6/106=0.1$$ mole

$$106g$$ contain atoms of oxygen $$=3\times 6.022\times { 10 }^{ 23 }$$

$$10.6$$ contain atoms $$=0.1\times 3\times 6.022\times { 10 }^{ 23 }$$
$$=1.8\times { 10 }^{ 23 }\ atoms$$
Question 9
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Which of the following is correct increasing order of molecular mass?

A
$${ H }_{ 2 }O<HCl<{ H }_{ 2 }S<{ CO }_{ 2 }$$

B
$$H_2O<H_2S<HCl<CO_2$$

C
$${ H }_{ 2 }O<CO_2<<HCl<{ H }_{ 2 }S$$

D
$$HCl<{ H }_{ 2 }S<{ H }_{ 2 }O<{ CO }_{ 2 }$$

Solution

Atomic mass of $$H=1$$

Atomic mass of $$O=16$$

Atomic mass of $$Cl=35.5$$

Atomic mass of $$C=12$$

Atomic mass of $$S=32$$

Molecular mass of the given compounds are-

$$H_2O=2+16=18$$

$$HCl=1+35.5=36.5$$

$$H_2S=2+32=34$$

$$CO_2=12+32=44$$

Hence, the correct increasing order for the molecular mass is,
$$H_2O<H_2S<HCl<CO_2$$
Question 10
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If $$x \ g$$ of an element $$A$$ contains $$\cfrac y{80}$$ atoms and $$2x \ g$$ of an element $$B$$ contains $$\cfrac y{40}$$ atoms then the ratio of atomic weights of two elements $$A$$ and $$B$$ is:

A
$$1:1$$

B
$$1:4$$

C
$$4:1$$

D
$$2:1$$

Solution

Let the atomic mass of $$A$$ be $$M_A$$ and $$B$$ be $$M_B$$
Given, $$\cfrac {x}{M_A} \times N_A=\cfrac y{80}....(i)$$ ($$N_A$$ is avogadro number)
$$\cfrac {2x}{M_B} \times N_A=\cfrac {y}{40}.....(ii)$$
$$(i)/(ii) \Rightarrow \cfrac {M_A}{M_B}=1:1$$

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Re-Attempt Weekly Quiz Competition

Atoms and Molecules Test - 59

Result

Atoms and Molecules Test - 59

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SHARING IS CARING

Question 1

$$94.117$$

$$1884$$

$$941.7$$

$$976$$

Solution

Question 2

$$1.505\times 10^{23}$$

$$3.01\times 10^{23}$$

$$6.02\times 10^{23}$$

none of these

Solution

Question 3

$$3.1\times 10^5$$

$$3.1$$

$$31$$

None of these

Question 4

the ratio of chemical species to each other in a balanced equation

the ratio of elements to each other in a compound

the definition of mass in units of grams

the mass of one mole of carbon

Solution

Question 5

$$3.42\times 10^4$$ amu

$$3.72\times 10^4$$ amu

$$3.6\times 10^4$$ amu

None of these

Solution

Question 6

$$400g$$

$$800g$$

$$200g$$

$$100g$$

Solution

Question 7

$$3.1 \times 10^{25}$$

$$6.2 \times 10^{25}$$

$$26$$

$$206$$

Solution

Question 8

$$6.02 \times 10^{22}$$

$$12.04 \times 10^{22}$$

$$1.806 \times 10^{23}$$

$$31.8$$

Solution

Question 9

$${ H }_{ 2 }O<HCl<{ H }_{ 2 }S<{ CO }_{ 2 }$$

$$H_2O<H_2S<HCl<CO_2$$

$${ H }_{ 2 }O<CO_2<<HCl<{ H }_{ 2 }S$$

$$HCl<{ H }_{ 2 }S<{ H }_{ 2 }O<{ CO }_{ 2 }$$

Solution

Question 10

$$1:1$$

$$1:4$$

$$4:1$$

$$2:1$$

Solution

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