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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Question 1

The composition of a sample of wustite is Fe0.95O1.00. Fe_{0.95} O_{1.00}. Fe0.95​O1.00​. What is the percentage of Fe(III) in the sample?

10.52%

3.46%

13.84%

93.08%

Solution

Question 2

Mg,5.8 gMg, 5.8\ gMg,5.8 g

Mg,0.58 gMg, 0.58\ gMg,0.58 g

O2,0.24 gO_{2}, 0.24\ gO2​,0.24 g

O2,2.4 gO_{2}, 2.4\ gO2​,2.4 g

Solution

Question 3

20.2420.2420.24 %

26.6426.6426.64 %

46.1946.1946.19 %

7.867.867.86 %

Question 4

2.0%2.0\%2.0%

6.4%6.4\%6.4%

10.4%10.4\%10.4%

2.4%2.4\%2.4%

Solution

Question 5

1,21, 21,2 and 333 are correct

111 and 222 only are correct

222 and 333 only are correct

111 only is correct

Solution

Question 6

222

0.020.020.02

202020

444

Solution

Question 7

Diamond inclusion

Fluid inclusion

Natural inclusion

Crystal inclusion

Question 8

66.6766.6766.67

24.0224.0224.02

72.0572.0572.05

69.6269.6269.62

Solution

Question 9

66.6766.6766.67

757575

707070

808080

Solution

Question 10

404040

58.858.858.8

76.4776.4776.47

17.6517.6517.65

Solution

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

$Mg, 5.8\ g$

$Mg, 0.58\ g$

$O_{2}, 0.24\ g$

$O_{2}, 2.4\ g$

$20.24$ %

$26.64$ %

$46.19$ %

$7.86$ %

$2.0\%$

$6.4\%$

$10.4\%$

$2.4\%$

$1, 2$ and $3$ are correct

$1$ and $2$ only are correct

$2$ and $3$ only are correct

$1$ only is correct

$2$

$0.02$

$20$

$4$

$66.67$

$24.02$

$72.05$

$69.62$

$66.67$

$75$

$70$

$80$

$40$

$58.8$

$76.47$

$17.65$