Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 62

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Weekly Quiz Competition

Question 1
1 / -0

The composition of a sample of wustite is $$ Fe_{0.95} O_{1.00}. $$ What is the percentage of Fe(III) in the sample?

A
10.52%

B
3.46%

C
13.84%

D
93.08%

Solution

The composition of a sample of wustite is $$Fe_{0.95}O$$. Formula of ferric oxide is $$FeO$$. So,number of $$Fe^{+2}$$ ion missing=0.05
Each $$Fe^{+2}$$ ion contains +2 charge. So total charge missing =2 $$\times$$0.05=0.10.
To maintain electrical neutrality the 0.10 positive charge is compensated by $$Fe^{+3}$$ ions.
Replacement of one ferrous ion by one ferric ion increases +1 charge. So no of ferric ions required to compensate the 0.10 charge is 0.10.
So 0.95 $$Fe^{+2}$$ ions require 0.1 $$Fe^{+3}$$ ions
So, in 100 Fe atoms, percentage of $$Fe^{+3}$$ is = $$(.1/.95)\times 100$$=10.52
Question 2
1 / -0

$$1.0\ g$$ of $$Mg$$ is burnt with $$0.28\ g$$ of $$O_{2}$$ in a closed vessel. Which reactant is left in excess and how much?

A
$$Mg, 5.8\ g$$

B
$$Mg, 0.58\ g$$

C
$$O_{2}, 0.24\ g$$

D
$$O_{2}, 2.4\ g$$

Solution

$$2Mg + O_{2}\rightarrow 2MgO$$

$$\therefore 1\ mole$$ of $$O_{2}$$ reacts with $$2$$ moles of $$Mg$$

$$\because$$ Moles of $$O_{2} = \dfrac {0.28}{32} = 0.00875$$

$$0.00875\ mole$$ of $$O_{2}$$ reacts with $$\dfrac {2}{1} \times 0.00875\ moles$$ of $$Mg$$

$$= 0.0715\ Moles$$ of $$Mg$$

Hence, mass of magnesium that reacts $$= moles\times molar\ mass$$

$$= 0.0715\times 24 = 0.42\ g$$

That means, out of the $$1\ g$$ of $$Mg$$, only $$0.42\ g$$ is used.

Therefore magnesium is in excess by $$(1 - 0.42) = 0.58\ g$$.

Hence, option $$B$$ is correct.
Question 3
1 / -0

Percentage of oxygen in urea is about :

A
$$20.24$$ %

B
$$26.64$$ %

C
$$46.19$$ %

D
$$7.86$$ %
Question 4
1 / -0

A sample of clay contains $$50\%$$ silica and $$10\%$$ water. The sample is partially dried by which it loses $$8\ g$$ water. If the percentage of silica in the partially dried clay is $$52$$, what is the percentage of water in the partially dried clay?

A
$$2.0\%$$

B
$$6.4\%$$

C
$$10.4\%$$

D
$$2.4\%$$

Solution

Let weight of clay is $$x\ gm$$
$$\therefore \ $$ Weight of silica $$=\dfrac {50}{100}\times x$$

$$=0.5\ x\ gm$$

$$\therefore \ $$ Weight of water $$=\dfrac {10}{100}\times x=0.1\ x\ gm$$
Now,
$$8\ gm$$ water is reduced
$$\therefore \ $$ weight of clay $$=(x-8)\ gm$$

Weight of silica $$=\dfrac {52}{100}\times (x-8)\ gm$$

$$=(0.52 \ x-4.16)\ gm$$

Now, Weight of silica in both case are equal.
$$\therefore \ 0.5\ x=0.52\ x-4.16$$

$$0.02\ x=4.16$$

$$\therefore \ x=208\ gm\ =$$ weight of clay initially.

$$\therefore \ $$ weight of water $$=0.1\ x=20.8\ gm$$

Now, $$8\ g$$ water is reduced.

$$\therefore \ $$ Weight of water $$=20.8-8\ x =12.8\ gm$$

Weight of clay $$=208-8\ gm=200\ gm$$

$$\therefore \ \%$$ of water $$=\dfrac {12.8}{200}\times 100=6.4\%$$
Question 5
1 / -0

Compound Q contains $$40\%$$ carbon by mass.
What could Q be?
$$1$$. glucose, $$C_6H_{12}O_6$$
$$2$$. starch, $$(C_6H_{10}O_5)_n$$
$$3$$. sucrose, $$C_{12}H_{22}O_{11}$$.

A
$$1, 2$$ and $$3$$ are correct

B
$$1$$ and $$2$$ only are correct

C
$$2$$ and $$3$$ only are correct

D
$$1$$ only is correct

Solution

$$\text{percentage mass of C}=\dfrac{\text{mass of C X 100%}}{\text{total mass}}$$

$$\text{1. percentage of C in glucose=40%}$$
$$\text{2. percentage of C in starch=44%}$$
$$\text{3. percentage of C in sucrose=42%}$$
Question 6
1 / -0

A mixture of $$He(4)$$ and $$Ne(20)$$ in a $$5-$$litre flask at $$300\ K$$ and $$1$$ atm weighs $$4\ g$$. The mole $$\%$$ of $$He$$ is

A
$$2$$

B
$$0.02$$

C
$$20$$

D
$$4$$

Solution

Using $$PV=nRT$$

$$n= \dfrac{PV}{RT} = \dfrac{1 \times 5}{0.082}{300} =0.203 \ moles $$

Let the amount of $$He$$ in the mixture be $$x$$ , thus amount of $$Ne$$ = $$4-x$$ g

Total moles = 0.203
$$\therefore \dfrac{x}{4} + \dfrac{4-x}{20} = 0.203$$

$$x= 0.012 \ g$$

Thus moles of $$He$$ in the mixture is $$\dfrac{0.012}{4} = 0.003$$

which is 2% moles of the mixture.
Question 7
1 / -0

What is involved in $$CO_2$$ and $$CH_4$$ inclusion?

A
Diamond inclusion

B
Fluid inclusion

C
Natural inclusion

D
Crystal inclusion
Question 8
1 / -0

A certain mixture of $$MnO$$ and $$MNO_2$$ contains $$66.6\ mol$$ per cent of $$MnO_2$$. What is the approximate mass percent of $$Mn$$ in it? $$(Mn=55)$$

A
$$66.67$$

B
$$24.02$$

C
$$72.05$$

D
$$69.62$$

Solution

Man of $$MnO=66.67\times 70.93$$
$$=4728.9\ g$$
man of $$MnO_2=33.33\times 86.93$$
$$=2897.37$$
Total man $$=4728.9+2897.37$$
$$=7626.2\ g$$
man of $$Mn=55\times 100=5500$$
$$\%Mn=\dfrac {5500}{7626.2}\times 100$$
$$=72.11\%$$
Question 9
1 / -0

A sample of impure cuprous oxide contains $$66.67\%$$ copper, by ,mass. What is the percentage of pure $$Cu_2O$$ in the sample? $$(Cu=63.5)$$

A
$$66.67$$

B
$$75$$

C
$$70$$

D
$$80$$

Solution

$$100\ gm$$ of sample contains $$=66.6\ gm$$
mole of $$Cu=\dfrac {66.6}{63.6}=1.05\ mole$$
$$1$$ mole of oxygen in $$2$$ mole $$Cu$$
$$2$$ mole of $$Cu$$ is with $$1$$ mole oxygen
$$1.05$$ mole contains $$\dfrac {1}{2}\times 1.05$$
weight of oxygen $$=52\times 16=84$$
pure $$Cu_2O=66.6+8.42$$
$$=75$$
Question 10
1 / -0

Directions For Questions

A sample of hydrogen fluoride gas (only $$HF$$ molecules) is collected in a vessel and left for some time. Then a constant molar mass of the sample is experimentally determined as $$34\ g/mole$$. Assume that this abnormal molar mass due to dimerization as well as trimerization of some $$HF$$ molecules (no molecules in any other polymeric fomrs) and the mole ratio of monomeric and trimeric from of hydrogen fluoride molecules present is $$4:1$$

...view full instructions

What percentage of hydrogen fluoride molecules is trimerized?

A
$$40$$

B
$$58.8$$

C
$$76.47$$

D
$$17.65$$

Solution

Consider mole of $$HE=4$$
$$H_{3}F_{3}=1$$ then $$HF$$ in it $$=1$$ molecules
$$H_{2}F_{2}=y$$ then $$HF$$ in it $$=2$$ molecules
Total mole $$=4+1+y=5+y$$
$$X_{HE}=\dfrac{4}{5+0}, X_{H_{2}F}=\dfrac{y}{5+y}, X_{H_{3}F_{3}}=\dfrac{1}{5+y}$$
$$20\left(\dfrac{y}{5+y}\right)+40\left(\dfrac{4}{5+y}\right)+60\left(\dfrac{1}{5+y}\right)=34$$
$$80+40y+60=170+94 y$$
$$40y-34y=170-140$$
$$6y=30$$
$$y=5$$
Total molecular $$m$$ monomer $$=4$$
dimmer $$=5\times 2=10$$
trimer $$=1\times 3=3$$
Total molecules $$=4+10+3=17$$

$$(5)\ \%$$ trimer $$=\dfrac{3}{17}\times 10=17.64\%$$ Option $$(d)$$

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Some Basic Concepts of Chemistry Test - 62

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Some Basic Concepts of Chemistry Test - 62

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Question 1

The composition of a sample of wustite is $$ Fe_{0.95} O_{1.00}. $$ What is the percentage of Fe(III) in the sample?

10.52%

3.46%

13.84%

93.08%

Solution

Question 2

$$Mg, 5.8\ g$$

$$Mg, 0.58\ g$$

$$O_{2}, 0.24\ g$$

$$O_{2}, 2.4\ g$$

Solution

Question 3

$$20.24$$ %

$$26.64$$ %

$$46.19$$ %

$$7.86$$ %

Question 4

$$2.0\%$$

$$6.4\%$$

$$10.4\%$$

$$2.4\%$$

Solution

Question 5

$$1, 2$$ and $$3$$ are correct

$$1$$ and $$2$$ only are correct

$$2$$ and $$3$$ only are correct

$$1$$ only is correct

Solution

Question 6

$$2$$

$$0.02$$

$$20$$

$$4$$

Solution

Question 7

Diamond inclusion

Fluid inclusion

Natural inclusion

Crystal inclusion

Question 8

$$66.67$$

$$24.02$$

$$72.05$$

$$69.62$$

Solution

Question 9

$$66.67$$

$$75$$

$$70$$

$$80$$

Solution

Question 10

$$40$$

$$58.8$$

$$76.47$$

$$17.65$$

Solution

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