Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 46

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Question 1
1 / -0

One mole of any substance contains $$6.022\times 10^{23}$$ atoms/molecules. Number of molecules of $$H_2SO_4$$ present in $$100$$mL of $$0.02$$M $$H_2SO_4$$ solution is __________.

A
$$12.044\times 10^{20}$$ molecules

B
$$6.022\times 10^{23}$$ molecules

C
$$1\times 10^{23}$$ molecules

D
$$12.044\times 10^{23}$$ molecules

Solution

Moles of $$H_2SO_4=$$ Molarity of $$H_2SO_4\times $$ Volume of solution $$(L)$$

$$=0.02\times 0.1$$

$$=2\times 10^{-3}$$ moles

No. of $$H_2SO_4$$ molecules $$=2\times 10^{-3}\times 6.022\times 10^{23}=12.044\times 10^{20}$$ molecules

Therefore, the correct option is A.
Question 2
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The weight of mixture containing HCl and $$H_2SO_4$$ is $$0.1$$g on treatment with an excess of an $$AgNO_3$$ solution, reacted with this acid mixture gives $$0.1435$$g of AgCl, weight percentage of $$H_2SO_4$$ in mixture is:

A
$$36.5$$

B
$$63.5$$

C
$$50$$

D
$$80$$

Solution

Let mass of $$HCl$$ be $$x$$
$$\therefore$$ Mass of $$H_2SO_4$$ is $$0.1-x$$
Moles of $$HCl=\cfrac{x}{36.5}$$
$$HCl+AgNO_3\longrightarrow AgCl\downarrow +HNO_3$$
Moles of $$HCl=$$ moles of $$AgCl$$
Moles of $$AgCl=\cfrac{0.1435}{143.3}=10^{-3}$$ moles
$$\therefore \cfrac{x}{36.5}=10^{-3}$$
$$x=36.5\times 10^{-3}$$
Weight of $$HCl=0.0365$$ $$g$$
Weight of $$H_2SO_4=0.0635$$
$$\%$$ weight of $$H_2SO_4=\cfrac{0.0635}{0.1}\times 100=63.5\%$$
Question 3
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The percentage composition of $$Cr$$ in $${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$$ is (Given molar mass of $${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }=294.0g$$ $$mol$$)

A
$$60.84$$

B
$$28.54$$

C
$$51.3$$

D
$$35.37$$

Solution

Given,
$$K_2Cr_2O_7$$
Molar mass of $$K_2Cr_2O_7=294$$ g/mol$$\longrightarrow (1)$$
Mass percentage=$$\cfrac {Mass\quad of\quad required\quad element\quad in\quad total\quad sample}{Mass\quad of\quad total\quad sample}\times 100$$
or mass percentage means, the amount of required element present in $$100g$$ of sample.
From, (1) $$1$$ mol i.e; $$294g$$ $$K_2Cr_2O_7$$ contains $$78g$$ $$K$$, $$104g$$ $$Cr$$ & $$112$$ $$O$$
Now, if $$294g$$ $$K_2Cr_2O_7$$ contains $$104g$$ $$Cr$$ then $$100g$$ sample of $$K_2Cr_2O_7$$ would contain:
$$\cfrac {100\times104}{294}=35.374g$$ of $$Cr$$
$$\therefore$$ Mass percentage of $$Cr$$ in $$K_2Cr_2O_7=35.37$$%
Question 4
1 / -0

Calculate the silver ion concentration in $$0.2\ M$$ solution as $$Ag(NH_{3})_{2}NO_{3} \cdot K_{silver} = 32\times 10^{-9}$$.

A
$$4\times 10^{-2}M$$

B
$$2\times 10^{-3}M$$

C
$$3\times 10^{-3}M$$

D
$$5\times 10^{-2}M$$

Solution

The reaction involved is:
$$[Ag(NH_3)_2]NO_3+H^+\longrightarrow Ag^++2NH_3+H^++{NO_3}^-$$
$$\Rightarrow K_p=\cfrac {[Ag][NH_3]^2[{NO_3}^-]}{[Ag(NH_3)_2]NO_3}$$
$$\Rightarrow 32\times 10^{-9}=\cfrac {[Ag]^4}{0.2}$$ $$[\because K_{og}=32\times 10^{-9}$$ given]
[complex]=$$0.2M$$ given
$$\Rightarrow Ag= 2\sqrt {2}\times 10^ {-2.5}$$
$$=2.8\times 10^{-2.5}\approx 3\times 10^{-3} M$$
Question 5
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In carbon disulphide $$(CS_{4})$$, the mass of sulphur in combination with $$3.0\ g$$ of carbon is

A
$$4.0\ g$$

B
$$6.0\ g$$

C
$$64.0\ g$$

D
$$16.0\ g$$

Solution

Given, Carbon disulphide $$CS_2$$
Molar mass of $$CS_2=12g+32\times 2g$$
$$=12+64g=76g$$
i.e; $$12g$$ of $$C$$ reacts with $$64g$$ of $$S$$ to form $$76g$$ $$CS_2$$
Also, in $$12g$$ of $$C$$ forms $$76g$$ $$CS_2$$
$$\Rightarrow 3.0g$$ of $$C$$ forms $$\cfrac {76\times3}{12}g$$ of $$CS_2$$
$$\Rightarrow 19g$$ of $$CS_2$$ is formed.
Also,
If $$76g$$ $$CS_2$$ has $$64g$$ of $$S$$
then, $$19g$$ $$CS_2$$ will have $$\cfrac {19\times 64}{76}g$$ of $$S$$
$$\Rightarrow 16.0g$$ of $$S$$ is present.
Question 6
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A solution containing 4.2 g of KOH and $$Ca(OH)_2$$ is neutralized by an acid. If it consumes 0.1 equivalent of acid, the % of KOH in the original sample is:

A
15%

B
25%

C
35%

D
45%

Solution

Let the amount of KOH in the solution be x, then amount of $$CaOH_2 $$ is (4.2-x)
Equivalent mass of KOH = $$\dfrac{56}{1} = 56 $$

Equivalent mass of $$ Ca(OH)_2 =\dfrac{74}{2} = 37 $$

Gram equivalents of KOH + Gram equivalents of
$$ Ca(OH)_2$$ = Gram equivalents of the acid

$$\Rightarrow \dfrac{x}{56}+\dfrac{4.2-x}{37}=0.1$$

$$\Rightarrow 37x + 235.2 - 56x = 207.2$$

$$\Rightarrow 235.2-207.2=56x-37x $$

$$\Rightarrow 28 = 19x $$

$$\Rightarrow x=\dfrac{28}{19} = 1.47 $$
Therefore, amount of KOH in the solution = 1.47
Percentage of KOH = $$\dfrac{1.47}{42} \times 100 = 35 $$
Hence, Option C is correct.
Question 7
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Sodium bicarbonate, $$NaHCO_{3}$$, can be purified by dissolving it in hot water $$(60^{\circ}C)$$, filtering to remove insoluble impurities, cooling to $$0^{\circ}C$$ to precipitate solid $$NaHCO_{3}$$, and then filtering to remove the solid, leaving soluble impurities in solution. Any $$NaHCO_{3}$$ that remains in the solution is not recovered. The solubility of $$NaHCO_{3}$$ in hot water at $$60^{\circ}C$$ is $$164\ g/litre$$ and is $$69\ g/litre$$ in cold water at $$0^{\circ}C$$. What is the percentage yield of $$NaHCO_{3}$$, when it is purified by this method?

A
$$55.34$$%

B
$$42.07$$%

C
$$69$$%

D
$$31$$%

Solution

During crystallization $$5\%$$ loss in acceptable expected yied $$=164-69=95\ g$$
$$5\%=\dfrac {x}{95}\times 100$$
$$x=\dfrac {5\times 95}{100}=4.75\ g$$
actual yield will get $$=95-4.75=90.25\ g$$
$$\%\ NaHCO_3$$ recverable $$=\dfrac {90.25}{164}\times 100$$
$$=55.03\%$$ option $$(a)$$
Question 8
1 / -0

$$H_3AO_4$$ has 31.6% of "A" by mass. The atomic mass of the atom "A" is:

A
17

B
31

C
35.5

D
40

Solution

Let the atomic mass of "A" be $$x$$.

Given,

31.6% of $$H_3AO_4=x$$

$$\Rightarrow \dfrac{31.6}{100}(1\times3+x+16\times4)=x$$

$$\Rightarrow 67+x=\dfrac{100}{31.6}x$$

$$\Rightarrow 67=\dfrac{68.4}{31.6}x$$

$$\Rightarrow x=31$$

Hence, option $$B$$ is correct.
Question 9
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$$\text{0.80 g}$$ of impure $$(NH_4)_2SO_4$$ was boiled with $$100 ml$$. of a $$\text{0.2 N NaOH}$$ solution till all the $$NH_3(g)$$ evolved.The remaining solution was diluted to $$\text{250 mL.25 mL}$$ of this solution was neutralized using $$5 mL$$ of a $$\text{0.2 n H_2SO_4}$$ solution.The percentage purity of the $$(NH_4)_2SO_4$$ sample is

A
$$82.5$$

B
$$72.5$$

C
$$62.5$$

D
$$17.5$$

Solution

Total m-eq. of $$NaOH $$ taken $$ =20$$
m-eq. of $$H_2SO_4 $$ = m-eq. of $$NaOH $$ reacted
$$ = \dfrac{5 \times 0.2}{25} \times 250 = 10 $$

m-eq. of $$NaOH $$ reacted $$ = 20 - 10 = 10 $$
$$ (NH_4)_2SO_4 + 2NaOH $$
$$ \Rightarrow Na_2SO_4 + 2NH_3 + 2H_2O $$
m-moles of $$(NH_4)_2SO_4 $$ reacted $$ = 5 $$
wt. of $$(NH_4)_2SO_4 $$ in sample
$$ = \dfrac{0.66}{0.80} \times 100 = 82.5 $$
Question 10
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Two organic compounds $$\underline {R}$$ and $$\underline {S}$$, both containing only $$C$$ and $$H$$ yield on analysis, the same percentage composition by weight.
$$C = (12/ 13) \times 100$$% and $$H = (1/13)\times 100$$%.
$$\underline {R}$$ decolourises $$Br_{2}$$ - water but $$\underline {S}$$ does not.

A
$$C_{2}H_{2}, C_{6}H_{6}$$

B
$$C_{6}H_{6}, C_{2}H_{2}$$

C
$$C_{2}H_{4}, C_{2}H_{6}$$

D
$$C_{2}H_{2}, C_{3}H_{8}$$

Solution

According to question the compound have equal % composition of mole. So, component carbon$$(C)$$ and the component $$H$$ have same in number.
$$\rightarrow \quad C_n H_n$$ type of compound
Now,
According to bromine water reaction see in fig.
But as this is question of physical chemistry, we know the $$C_n H_n$$ type of molecules should be obtained but in options only $$(A)$$ option has only $$C_n H_n$$ like atoms.
So $$(A)$$ is correct answer.

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Some Basic Concepts of Chemistry Test - 46

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Some Basic Concepts of Chemistry Test - 46

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Question 1

$$12.044\times 10^{20}$$ molecules

$$6.022\times 10^{23}$$ molecules

$$1\times 10^{23}$$ molecules

$$12.044\times 10^{23}$$ molecules

Solution

Question 2

$$36.5$$

$$63.5$$

$$50$$

$$80$$

Solution

Question 3

$$60.84$$

$$28.54$$

$$51.3$$

$$35.37$$

Solution

Question 4

$$4\times 10^{-2}M$$

$$2\times 10^{-3}M$$

$$3\times 10^{-3}M$$

$$5\times 10^{-2}M$$

Solution

Question 5

$$4.0\ g$$

$$6.0\ g$$

$$64.0\ g$$

$$16.0\ g$$

Solution

Question 6

15%

25%

35%

45%

Solution

Question 7

$$55.34$$%

$$42.07$$%

$$69$$%

$$31$$%

Solution

Question 8

17

31

35.5

40

Solution

Question 9

$$82.5$$

$$72.5$$

$$62.5$$

$$17.5$$

Solution

Question 10

$$C_{2}H_{2}, C_{6}H_{6}$$

$$C_{6}H_{6}, C_{2}H_{2}$$

$$C_{2}H_{4}, C_{2}H_{6}$$

$$C_{2}H_{2}, C_{3}H_{8}$$

Solution

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