Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 73

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Weekly Quiz Competition

Question 1
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What is the mass per cent of oxygen in ethanol?

A
$$52.14\%$$

B
$$13.13\%$$

C
$$16\%$$

D
$$34.73\%$$

Solution

Ethanol is $$C_2H_6O$$
molar mass of ethanol =46 g
Mass % of oxygen =$$100 \times \dfrac{(mass \ of\ oxygen\ in\ the\ molecule)}{(molar\ mass\ of\ molecule)}$$
$$\therefore$$ m% of oxygen $$=100 \times (16/46)=34.78$$%
Hence, option $$D$$ is correct.
Question 2
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0.2 g of an organic compound contains $$C,\ H,\ O$$ on combustion, it yields 0.15 g of $$CO_2$$ and 0.12 g $$H_2O$$. The percentage of $$C,\ H$$ and $$O$$ respectively is:

A
$$C=15$$ %, $$H= 20$$ %, $$O=65$$ %

B
$$C=10$$ %, $$H=8.2$$ %, $$O=81.8$$%

C
$$C=12.2$$ %. $$H=8.8$$ %, $$O=79$$ %

D
$$C=20$$ %, $$H=6.66$$ %, $$O=73.34$$ %

Solution

Weight of organic compound = 0.2g
Weight of $$C{O}_{2}$$ = 0.15g
Weight of $${H}_{2}O$$ = 0.12g
Molecular weight of $$C{O}_{2}$$ = 44
Molecular weight of $${H}_{2}O$$ = 18
Atomic weight of carbon = 12u
Atomic weight of $${H}_{2}$$ = 2u
$$\text{Percentage of carbon} = \cfrac{\text{Atomic weight of carbon}}{\text{Molecular weight of }C{O}_{2}} \times \cfrac{\text{Weight of }C{O}_{2}}{Weight of organic compound} \times 100$$
$$\text{Percentage of carbon} = \cfrac{12}{44} \times \cfrac{0.15}{0.2} \times 100 = 20.45 \% \approx 20 \%$$
$$\text{Percentage of Hydrogen} = \cfrac{\text{Atomic weight of Hydrogen}}{\text{Molecular weight of }{H}_{2}O} \times \cfrac{\text{Weight of }{H}_{2}O}{Weight of organic compound} \times 100$$
$$\text{Percentage of Hydrogen} = \cfrac{2}{18} \times \cfrac{0.12}{0.2} \times 100 = 6.66 \%$$
Percentage of oxygen = 100 - (Percentage of carbon + Percentage of hydrogen) = 100 - (20 + 6.66) = 100 - 26.66 = 73.34 $$\%$$
Question 3
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$$2$$g of brass containing $$Cu$$ and $$Zn$$ only reacts with $$3$$M $$HNO_3$$ solution. Following are the reactions taking place.

$$Cu(s)+HNO_3(aq)\rightarrow Cu^{2+}(aq)+NO_2(g)+H_2O(l)$$

$$Zn(s)+H^+(aq)+{NO}^-_3(aq)\rightarrow NH^+_4(aq)+Zn^{2+}(aq)+H_2O(l)$$

The liberated $$NO_2(g)$$ was found to be $$1.04$$L at $$25^o$$C and $$1$$ atm $$[Cu =63.5, Zn=65.4]$$.
How many grams of $$NH_4NO_3$$ will be obtained in the reaction?

A
$$0.405$$g

B
$$0.0428$$g

C
$$0.2018$$g

D
$$0.358$$g

Solution

$$ \text { Balance equation } $$
$$ \text { Cu }+ 4 \mathrm{H} \mathrm{N} \mathrm{O}_{3} \longrightarrow \mathrm{Cu}\left(\mathrm{N} \mathrm{O}_{3}\right)_{2}+2 \mathrm{~N} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$

$$ P V= n R T $$
$$ \begin{array}{l} V=1.04 L \\ P=atm \\ T=25^{\circ} C=290 K \\ R=0.0821 \text { . atm } / \mathrm{mol}^{-1} \mathrm{k}^{-1} \end{array} $$

$$ \begin{aligned} \text { moles of } \mathrm{NO}_{2} &=\dfrac{1 \times 1.04}{0.0821 \times 298} \\ &=0.0425 \end{aligned} $$

$$ \begin{aligned} \text { moles of } C u=\dfrac{1}{2} \times & \text { mole of } NO_2 \\ =& \dfrac{1}{2} \times 0.0425 \\ &=0.02125 \end{aligned} $$

$$ \begin{aligned} \text { weight of } Cu &=\operatorname{mole} x \text { molecular weight } \\ &=0.02125 \times 63.5 . \\ &=1.3496 g \end{aligned} $$

$$ \begin{aligned} \text { weight of } Zn=& \text { weight of brass - weight of Cu} \\ =& 2-1.3496 \\ &=0.6504 \end{aligned} $$
$$ \text { more of } Z n=\dfrac{0.6504}{65.4} $$

$$ \text { moles of } \mathrm{N} \mathrm{H}_4\mathrm{~N}O_3 \mathrm{~L}=\dfrac{2}{1} \dfrac{1}{4} \times \text { mole of } Z n$$

$$ =\dfrac{1}{2} \times \dfrac{0.6504}{65.4}=2.6 \times 10^{-3} $$

$$ \begin{aligned} \text { weight of } NH_4NO_3 &=\text { mole } \times \text { molecular weight } NH_4NO_3 \\ &=2.6 \times 10^{-3} \times 80 \\ &=0.2018 \mathrm{~g} \end{aligned} $$
Question 4
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Calculate the mass percent (w/w) of sulphuric acid in a solution prepared by dissolving $$4$$ g of sulphur trioxide in a $$100$$ml sulphuric acid solution containing $$80$$ mass percent (w/w) of $$H_2SO_4$$ and having a density of $$1.96\ g/ml$$(molecular weight of $$H_2SO_4=98$$) Take reaction $$SO_3+H_2O\rightarrow H_2SO_4$$.

A
$$80.8\%$$

B
$$84\%$$

C
$$41.65\%$$

D
None of these

Solution

$$SO_3 + H_2O \rightarrow H_2SO_4$$
100 ml sulphuric acid solution
density = $$\dfrac{Mass}{volume}$$
$$1.96 \, g/ml = \dfrac{Mass}{100}$$
Mass $$= 1.96 g$$
Amount of water $$= 196 \times 20$$
$$= 39.2 g$$
Amount of $$H_2SO_4 = 156.8 g$$
$$SO_3 + H_2O \rightarrow H_2 SO_4$$
Molar mass of $$SO_3 = 32 + 16 \times 3 = 80$$
No. of mole = $$49$$
$$80 g = 0.05 mol$$
0.05 mol of sulphuric acid formed
Now total amount of sulphuric acid
$$= 0.05 \times 98 + 156.8$$
$$= 4.9 + 156.8$$
$$= 161.7$$
total amount of water $$= 39.2 - $$ total water consume
$$= 39.2 - 0.05 \times 18$$
$$=39.2 - 0.9 = 38.3$$
Mass percent (w/w) $$= \dfrac{161.7}{161.7 + 383} = \dfrac{161.7}{200} \times 100 \% $$
$$= 80.85\%$$
$$\approx 80.8\%$$
Option $$A$$ is correct.
Question 5
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Mole fraction of ethyl alcohol in aqueous ethyl alcohol $$(C_2H_5OH)$$ solution is $$0.25$$. Hence percentage of ethyl alcohol by weight is:

A
$$54\%$$

B
$$25\%$$

C
$$75\%$$

D
$$46\%$$

Solution

If ethanol is 0.25, then H$$_2$$O will be 0.75.
Mol wt of ethanol is 46.06 g/mol,
So, 0.25 mol=11.515g of ethanol

Mol wt of H$$_2$$O is 18.01528g/mol,
So, 0.75 mol=13.51g of H$$_2$$O

Total wt=25.02646 g

wt % of etanol is $$={\dfrac{ 11.515}{25.02646}} \times 100 = 46.011 \approx 46$$%

Hence, the correct option is $$D$$
Question 6
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32 g of a sample of $$FeSO_4$$.$$7H_2O$$ were dissolved in dilute sulphuric acid and water and its volume was made up to 1 litre. 25 mL of this solution required 20 mL of 0.02 M $$KMnO_4$$ solution for complete oxidation. Calculate the mass % of $$FeSO_4$$.$$7H_2O$$ in the sample.

A
34.75

B
69.5

C
89.5

D
None of these

Solution

$$Fe^{+2}\longrightarrow Fe^{+3}$$
$$n_{factor}=1$$
Comparing equivalents of $$FeSO_4\cdot 7H_2O$$ and $$KMnO_4$$
Let molarity of $$FeSO_4\cdot 7H_2O=M$$
$$25\times M\times 1=20\times 0.02\times 5$$
$$M=0.08$$ $$Molar$$
No. of moles of $$FeSO_4\times 7H_2O=0.08$$ $$moles$$ (Volume $$=1$$ Litre)
Mass of $$FeSO_4\cdot 7H_2O=0.08\times 278=22.24$$ $$gm$$
$$Mass\%=\cfrac{22.24}{32}\times 100=69.5\%$$
Question 7
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Carbon occurs in nature as a mixture of $$^{12}C$$ and $$^{13}C$$. The average atomic mass of carbon is 12.011. What is the % abundance of $$^{12}C$$ in nature?

A
88.9%

B
98.9%

C
89.9%

D
79.9%

Solution

Average atomic mass of an element existing in different isotopes is given by:
$$M_{ avg }=\frac { \sum _{ i=1 }^{ n }{ { M }_{ i }{ A }_{ i } } }{ \sum _{ i=1 }^{ n }{ A_{ i } } } $$
where
$$M_i=$$atomic mass of an isotope with relative abundance of $$A_i$$

Given:
$$M_{avg}=12.011\ u, M_1=12 u,A_1=a,M_2=13 u, A_2=1-a$$

On subtitutiing we get:

$$12.011=\frac { 12\times a+13\times (1-a) }{ a+(1-a) } $$

$$12.011=13-a$$

$$a=0.989$$

Thus relative abundance of $$^{12}C$$ is 98.9% and of $$^{13}C$$ is 1.1%
option B is correct
Question 8
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An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. What will be the masses of the carbon dioxide and water produces when 0.20g of this substance is subjected to the complete combustion?

A
0.69 g and 0.048g

B
0.506 g and 0.086g

C
0.345 g and 0.0024 g

D
0.91 g and 0.72 g

Solution

Given :
% of $$C = 69\%$$, % of $$H = 4.8 \%$$

$$44g$$ of $$CO_2$$ has $$12g $$ of C

$$1 g$$ of $$CO_2 = \dfrac{12}{44} g$$ of C

$$\% C = \dfrac{12}{44} \times \dfrac{w_c}{0.2} \times 100$$

$$\dfrac{44 \times 69 \times 0.2}{12 \times 100} = w_c$$

$$w_c = 0.506$$

$$18 g$$ of $$H_2 O$$ has $$2g$$ of H

$$1g $$ of $$H_2 O = \dfrac{2}{18} gH$$

$$w_{H_2} O$$ of $$H_2O$$ has $$\dfrac{2}{18} \times w_{H_2O} $$

$$\%$$ of $$H = \dfrac{2}{18} \times \dfrac{w_{H_2O}}{0.2} \times 100$$

$$4.8 = \dfrac{2}{18} \times \dfrac{w_{H_2O}}{0.2} \times 100$$

$$w_{H_2O} = 0.0864 g$$

Weight of Carbon and Hydrogen in the organic compound are $$0.506 g$$ and $$0.0864 g$$, respectively.
Question 9
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A mixture of $$NaHCO_3$$ and $$Na_2CO_3$$ weighs 950 gm. When the mixture is heated strongly in an open vessel, till constant weight, the mass of sample reduced by 155 gm. The correct information(s) is/ are ?

A
The mass percent of $$NaHCO_3$$ in initial mixture = 50%

B
The mole percent of $$Na_2CO_3$$ in initial mixture = 50%

C
Final sample contains only $$Na_2O$$

D
The loss in mass is only due to $$CO_2$$ gas evolved

Solution

$${ Na }_{ 2 }{ CO }_{ 3 }$$ is a strong base and does not dissociate on normal heating.
$$2NaH{ CO }_{ 3 }\longrightarrow { Na }_{ 2 }{ CO }_{ 3 }+{ CO }_{ 2 }+{ H }_{ 2 }O$$
No. of moles $${ CO }_{ 2 }=\dfrac { 0.950 }{ 22400 } =4.24\times { 10 }^{ -5 }$$
No. of moles of $${ NaHCO }_{ 3 }=4.24\times { 10 }^{ -5 }\times 22400=0.084\times { 10 }^{ +3 }=84g$$
Initial $${ NaHCO }_{ 3 }$$ mass $$=155g$$
$$\Rightarrow $$ Mass percent $$=\dfrac { 84 }{ 155 } \times 100$$% $$\approx 50$$%
Question 10
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P and Q are two elements which form $${ P }_{ 2 }{ Q }_{ 3 }$$ and $${ PQ }_{ 2 }$$. If 0.15 mole of $${ P }_{ 2 }{ Q }_{ 3 }$$ weight 15.9 g and 0.15mole of $${ PQ }_{ 2 }$$ weight 9.3 g. what are atomic weights of P and Q respectively?

A
18 and 26

B
26 and 26

C
26 and 18

D
18 and 18

Solution

Using- no. of mole $$=\cfrac{\text {weight}}{\text{molecular weight}}$$

$$P_2Q_3 \Rightarrow $$ $$2P + 3Q =15.9/0.15= 106$$....(1)

$$PQ_2 \Rightarrow $$ $$P + 2Q = 9.3/0.15=62$$........(2)

On solving eqn. (1) & (2) we get-

$$P =26, Q = 18$$

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 73

Result

Some Basic Concepts of Chemistry Test - 73

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SHARING IS CARING

Question 1

$$52.14\%$$

$$13.13\%$$

$$16\%$$

$$34.73\%$$

Solution

Question 2

$$C=15$$ %, $$H= 20$$ %, $$O=65$$ %

$$C=10$$ %, $$H=8.2$$ %, $$O=81.8$$%

$$C=12.2$$ %. $$H=8.8$$ %, $$O=79$$ %

$$C=20$$ %, $$H=6.66$$ %, $$O=73.34$$ %

Solution

Question 3

$$0.405$$g

$$0.0428$$g

$$0.2018$$g

$$0.358$$g

Solution

Question 4

$$80.8\%$$

$$84\%$$

$$41.65\%$$

None of these

Solution

Question 5

$$54\%$$

$$25\%$$

$$75\%$$

$$46\%$$

Solution

Question 6

34.75

69.5

89.5

None of these

Solution

Question 7

88.9%

98.9%

89.9%

79.9%

Solution

Question 8

0.69 g and 0.048g

0.506 g and 0.086g

0.345 g and 0.0024 g

0.91 g and 0.72 g

Solution

Question 9

The mass percent of $$NaHCO_3$$ in initial mixture = 50%

The mole percent of $$Na_2CO_3$$ in initial mixture = 50%

Final sample contains only $$Na_2O$$

The loss in mass is only due to $$CO_2$$ gas evolved

Solution

Question 10

18 and 26

26 and 26

26 and 18

18 and 18

Solution

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