Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 71

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Weekly Quiz Competition

Question 1
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Percentage purity of a sample of gold is $$85\%$$. How many atoms of gold are present in its $$1$$ gram sample? (Atomic mass of gold $$=197$$u).

A
$$3.0\times 10^{21}$$

B
$$2.6\times 10^{23}$$

C
$$2.6\times 10^{21}$$

D
$$4.5\times 10^{26}$$

Solution

Given,
$$85$$% pure gold sample
That means in $$100gm$$ sample $$85gm$$ gold is present
We are given with $$1gm$$ sample
$$\therefore$$ Mass of gold=$$85\times 10^{-2}$$
We know, $$1$$ mole of gold=$$197gm$$
$$197gm$$ gold=$$6.023\times 10^{23}$$ atoms of gold
$$\therefore 85\times 10^{-2}$$ gm gold contains $$\cfrac {85\times 10^{-2}\times 6.023\times 10^{23}}{197}$$ atoms
$$=2.598\times 10^{21}$$ atoms
$$=2.6\times 10^{21}$$ atoms
Question 2
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Directions For Questions

A $$10\ ml$$ mixture of $$N_{2}$$, a alkane and $$O_{2}$$ undergo combustion in Eudiometry tube. There was contraction of $$2\ ml$$, when residual gases are passed through $$KOH$$. To the remaining mixture comprising of only one gas excess $$H_{2}$$ was added and after combustion the gas produced is absorbed by water, causing a reduction in volume of $$8\ ml$$.

...view full instructions

Volume of $$O_{2}$$ remained after the first combustion?

A
$$4\ ml$$

B
$$2\ ml$$

C
$$0$$

D
$$8\ ml$$

Solution

Option C is correct
Question 3
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Calculate the mole percentage of $$CH_{3}OH$$ and $$H_{2}O$$ respectively in $$60$$% (by mass) aqueous solution of $$CH_{3}OH$$.

A
$$45.8, 54.2$$

B
$$54.2, 45.8$$

C
$$50, 50$$

D
$$60, 40$$

Solution

Given,
$$60$$% by mass solution of $$CH_3OH$$. That means in $$100gm$$ solution

$$60gm$$ $$CH_3OH$$ is dissolved in $$40gm$$ $$H_2O$$

$$\therefore$$ Molar mass of $$CH_3OH=32g/mol$$

Molar mass of $$H_2O=18g/mol$$

$$\Rightarrow$$ No. of moles of $$CH_3OH=\cfrac {60gm}{32g/mol}=1.875\ mol $$ $$..........(1)$$

No. of moles of $$H_2O=\cfrac {40gm}{18gm/mol}=2.222\ mol $$ $$..........(2)$$

Total no. of moles=$$1.875+2.222=4.097$$ moles $$..........(3)$$

$$\Rightarrow$$ Mole percentage= $$\cfrac {Moles\ of\ compound}{Total\ no.\ of\ moles}\times 100$$

Mole % of $$CH_3OH=\cfrac {1.875}{4.097}\times 100$$

$$=45.8$$% (From 1 & 3)

Mole % of $$H_2O=\cfrac {2.222}{4.097}\times 100$$
$$=54.23$$% (from 2,3)

Hence, the correct option is $$A$$
Question 4
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The vapour pressure of a solution of a non -volatile electrolyte (A) in a solvent (B) is $$95\%$$ of the vapour pressure of the solvent at the same temperature. If $$M_b=0.3 M_A$$, where $$M_B$$and $$M_A$$ are molecular weights of B and A respectively, the weight ratio of the solvent and solute are:

A
0.15

B
5.7

C
0.2

D
4.0

Solution

Given,
vapour pressure of solution having non volatile solute & solvent =$$35$$% of vapour pressure of pure solvent.
At constant temperature
Let, $$P^o$$ be vapour pressure of pure solvent
Let $$P_s$$ be vapour pressure of solution
$$\Rightarrow$$ If $$P^o=100$$ then $$P_s=95$$ $$[\because given]$$
as $$95$$% of $$100=95$$
From Relative lowering of vapour pressure , we know
$$\cfrac {P^o-P_s}{P^o}=x_{solute}$$ $$[ x_{solute}$$ is mole fraction of solute $$]$$
$$\Rightarrow \cfrac {5}{100}=x_{solute}$$
$$\Rightarrow x_{solute}=0.05 \longrightarrow (1)$$
Now, we know
$$x_solute=\cfrac {No.\quad of\quad moles\quad of\quad solute}{No.\quad of\quad moles\quad of\quad solvent}$$
$$\Rightarrow X_{solute}=\cfrac {n_A}{n_B}$$ $$[\because n_A <<< nB]$$
$$\Rightarrow X_{solute}= \cfrac {W_A}{M_A}\div \cfrac {W_B}{M_B}$$
$$\Rightarrow X_{solute}=\cfrac {W_A}{W_B}\times \cfrac {M_B}{M_A}$$
$$\Rightarrow 0.05=\cfrac {W_A}{W_B}\times 0.3$$ $$[\because M_B=0.3 M_A$$ given $$]$$
$$\Rightarrow \cfrac {W_A}{W_B}=\cfrac {0.05}{0.3}$$
$$=0.166\approx 0.15$$
Question 5
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A 5 mL (specific gravity 1.02) of chlorine water is treated with an excess of KI.The liberate iodine required 26 mL of $$ 0.15 \,M-Na_2SO_4$$ solution.Calculate the percentage of $$Cl_2$$ (by weight) the chloride water.Chloride water is a solution of free chlorine in water

A
$$0.388 \%$$

B
$$0.301 \%$$

C
$$0.307 \%$$

D
$$3.02 \%$$
Question 6
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$$0.3\ g$$ of an oxalate salt was dissolved in $$100\ mL$$ solution. The solution required $$90\ mL$$ of $$N/20\ KMnO_4$$ for complete oxidation.The % of oxalate ion in salt is:

A
$$33\%$$

B
$$66\%$$

C
$$70\%$$

D
$$40\%$$

Solution

Reaction taking between potassium permanganate and axalate-
$$MnO_4^- +C_2O_4^{2-} + H^+ \to Mn^{2+}+CO_2 +H_2O$$
Hence equivalent of $$KMnO_4$$ oxidizes $$2.5$$ equivalences of axalate
Moles of $$KMnO_4 =0.05\times \dfrac {0.09}{5}=9\times 10^{-4} .....(1)$$
Moles of oxalate that $$MKnO_4 $$ axidizes $$=9\times 10^{-4} \times 2.5$$
$$=22.5\times 10^{-4}$$ (from $$(1)$$)$$......(2)$$
Molecolar weight of oxalate $$=2\times 12+4\times 16=88$$
($$\because C_2 O_4$$, where $$C=12$$ & $$O=16$$)$$......(3)$$
Oxalate ion amount $$=22.5\times 10^{-4}\times 88 =0.198$$ from $$(2)$$ & $$(3)$$
Percentage of oxalate ion $$=\dfrac {0.198\times 100}{0.3}=66\%$$
Ans $$B)\ 66\%$$
Question 7
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How many moles of magnesium phosphate, $$Mg_3(PO_4)_2$$ will contain 0.25 mole of oxygen atoms ?

A
$$0.02$$

B
$$3.125 \times 10^{-2}$$

C
$$1.25 \times 10^{-2}$$

D
$$2.5 \times 10^{-2}$$

Solution

$$1$$ molecule of $$Mg_3(PO_4)_2$$ has $$8$$ atoms of Oxygen.

$$\Rightarrow 1$$ mole of $$Mg_3(PO_4)_2$$ has $$8$$ mole of Oxygen atoms.

Now, it is given $$0.25$$ mole of oxygen atoms
$$\therefore$$ Moles of $$Mg_3(PO_4)_2$$ having $$0.25$$ mole of $$O$$ atoms is $$\cfrac {0.25\times 1}{8}$$ moles
$$=0.03125$$ moles

$$=3.125\times 10^{-2}$$ moles

Hence, option B is correct.
Question 8
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A 1.85 g sample of an arsenic-containing pesticide was chemically converted to $$AsO_4^{3-}$$ (atomic mass of As = 74.9) and titrated with $$Pb^{2+}$$ to form $$Pb_3(AsO_4)_2$$. If 20 mL of 0.1 M $$PB^{2+}$$ is required to reach the equivalence point, the mass percentages of arsenic in the pesticide sample is closest to

A
8.1

B
2.3

C
5.4

D
3.6

Solution

Given, Mass of sample=$$1.85g$$
The reaction involved is
$$3Pb^{2+}+2AsO_4^{3-}\longrightarrow Pb_3(AsO_4)_2$$
Now, Volume of $$Pb^{2+}$$ consumed for neutralization=$$20ml$$
Molarity of $$Pb^{2+}=0.1M$$
$$\therefore$$ No. of moles of $$Pb^{2+}=0.1\times \cfrac {20}{1000}$$
$$=2\times 10^{-3}$$ moles
Now, from reaction it is clear that
$$3$$ moles $$Pb^{2+}$$ produce $$1$$ mole $$Pb_3(AsO_4)_2$$
We know $$1$$ mole $$Pb_3(AsO_4)_2$$ has $$1$$ mole $$As$$ in it.
$$\Rightarrow 3$$ moles $$Pb^{2+}$$ produce $$1$$ mole $$As$$
$$\therefore 2\times 10^{-3}$$ moles $$Pb^{2+}$$ will give $$\cfrac {2\times 10^{-3}}{3}$$ moles of $$As$$
$$\therefore$$ No. of moles of $$As$$ in the compound=$$0.66\times 10^{-3}$$ moles
Now,
We know, $$1$$ mole $$As=74g$$
$$\Rightarrow 0.66\times 10^{-3}$$ moles $$As=0.66\times 10^{-3}\times 74g$$
$$=48.84\times 10^{-3}g$$
$$=0.0488g$$
Now % calculation,
Mass% of $$As=\cfrac {Mass\quad of\quad As}{Mass\quad of\quad Sample}\times 100$$
$$=\cfrac {0.0488}{1.85}\times 100$$
$$=0.0263\times 100$$
$$=2.6$$% $$\approx 2.3$$%
Question 9
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A $$1.85\ g$$ sample of an arsenic-containing pesticide was chemically converted to $$AsO_{4}^{3-}$$ (atomic mass of $$As = 74.9)$$ and titrated with $$Pb^{2+}$$ to form $$Pb_{3}(AsO_{4})_{2}$$. If $$20\ mL$$ of $$0.1\ M\ Pb^{2+}$$ is required to reach the equivalence point, the mass percentage of arsenic in the pesticide sample is closest to:

A
$$8.1$$

B
$$2.3$$

C
$$5.4$$

D
$$3.6$$
Question 10
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4.0 g of a mixture of $$NaCl$$ and an unknown metal iodide $$MI_{2}$$ was dissolved in a water to form its aqueous solution. To this aqueous solution, The aqueous solution $$AgNO_{3}$$ was added gradually so that silver halides are precipitated. The precipitates were weighed at regular interval and the given curve for the mass of precipitate versus volume of $$AgNO_{3}$$ added was obtained. With the knowledge of the fact that halides are precipitated successively i.e. when one halide is precipitating , the other halide remains in the solution, answer the following questions:
(Molar mass of $$Ag = 108,\ I = 127,\ Na = 23$$).
Points $$P$$ and $$Q$$ are: $$P(20, 4.8),\ Q(37, 7.25)$$
What is the approximate mass percentage of $$MI_{2}$$?

A
25

B
40

C
60

D
75

Solution

Mass of $$\mathrm{AgCl}$$ (formed in $$2^{\text {nd }}$$ step $$)$$
$$=7.25-4.80=2.45 \mathrm{~g}$$
$$\because 143.5 \mathrm{~g} \mathrm{AgCl}$$ form $$58.5 \mathrm{~g}$$ Nacl
2.45 g $$\mathrm{AgCl}$$ form $$\frac{58.5}{143.5}\times 2.45$$
$$=1.0 \mathrm{~g} \mathrm{NaCl}$$
Mass of $$M I_{2}=4-1=3 g$$
$$\begin{aligned}\% \text { of } M I_{2} &=\frac{3}{4} \times 200 \\&=75 \%\end{aligned}$$
$$\therefore$$ (D) is correct coption.

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 71

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Some Basic Concepts of Chemistry Test - 71

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SHARING IS CARING

Question 1

$$3.0\times 10^{21}$$

$$2.6\times 10^{23}$$

$$2.6\times 10^{21}$$

$$4.5\times 10^{26}$$

Solution

Question 2

$$4\ ml$$

$$2\ ml$$

$$0$$

$$8\ ml$$

Solution

Question 3

$$45.8, 54.2$$

$$54.2, 45.8$$

$$50, 50$$

$$60, 40$$

Solution

Question 4

0.15

5.7

0.2

4.0

Solution

Question 5

$$0.388 \%$$

$$0.301 \%$$

$$0.307 \%$$

$$3.02 \%$$

Question 6

$$33\%$$

$$66\%$$

$$70\%$$

$$40\%$$

Solution

Question 7

$$0.02$$

$$3.125 \times 10^{-2}$$

$$1.25 \times 10^{-2}$$

$$2.5 \times 10^{-2}$$

Solution

Question 8

8.1

2.3

5.4

3.6

Solution

Question 9

$$8.1$$

$$2.3$$

$$5.4$$

$$3.6$$

Question 10

25

40

60

75

Solution

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