Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

A sample of municipal water contains one part of urea (molecular wt $$=60$$) per million parts of water by weight. The number of urea molecules in a drop of water of volume $$0.05\ ml$$ is

A
$$2.5\times 10^{14}$$

B
$$5\times 10^{14}$$

C
$$5\times 10^{13}$$

D
$$5\times 10^{15}$$

Solution

$$1 \ ppm = 1 \ mg/L = 10^{-3} g/L$$

Water sample contains $$1 \ ppm$$ urea concentration.

$$\therefore \ 10^{-3} \ g$$ of urea in $$ 1 \ L$$

$$x \ g$$ urea in $$0.05 \times 10^{-3} \ L$$

$$x= 0.05 \times 10^{-6} \ g$$

$$60 \ g$$ urea $$=6.023 \times 10^{23} \ molecules$$

$$0.05 \times 10^{-6} \ g = n \ molecules$$

$$n = \cfrac {6.023 \times 10^{23} \times 0.05 \times 10^{-6}}{60}$$

$$=0.005 \times 10^{17}$$

$$=5 \times 10^{14} \ molecules$$
Question 2
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If you are given Avogadro's number of atoms of a gas $$X$$. If half of the atoms are converted into $$X_{(g)}^+$$ by energy $$\Delta H$$. The IE of $$X$$ is :

A
$$\dfrac{2\Delta H}{N_A}$$

B
$$\dfrac{2N_A}{\Delta H}$$

C
$$\dfrac{\Delta H}{2N_A}$$

D
$$\dfrac{N_A}{\Delta H}$$

Solution

Given no. of atoms = Avogadro's no. of atoms = $${N}_{A} = 6.023 \times {10}^{23}$$
Given that $$\cfrac{{N}_{A}}{2}$$ atoms are ionized, i.e.,
Ionization energy of $$\cfrac{{N}_{A}}{2}$$ atoms of gas X = $$\Delta{H}$$
$$\therefore$$ Ionization energy of 1 atom of gas X = $$\cfrac{\Delta{H}}{\left( \cfrac{{N}_{A}}{2} \right)} = \cfrac{2. \Delta{H}}{{N}_{A}}$$
Hence, Ionisation energy of gas X is $$\cfrac{2. \Delta{H}}{{N}_{A}}$$.
Question 3
1 / -0

The percentage of pyridine $$(C_5H_5N)$$ that forms pyridinium ion $$(C_5H_5NH)$$ in a $$0.10$$M aqueous pyridine solution is:
[$$K_b$$ for $$C_5H_5N=1.7\times 10^{-9}$$]

A
$$0.77\%$$

B
$$1.6\%$$

C
$$0.0060\%$$

D
$$0.013\%$$

Solution

We know that

$$K_a = \cfrac {K_w}{K_b}= \cfrac {10^{-14}}{1.7 \times 10^{-9}}= 5.88 \times 10^{-6}$$

$$C_5H_5N+ H_2O \rightleftharpoons C_5H_5NH + C_5H_5NH^+ + OH^-$$

Initial conc: $$0.1$$ $$0$$ $$0$$ $$0$$

Final Conc: $$0.1-x$$ $$x$$ $$x$$

$$K_a = \cfrac {x^2}{0.1-x}$$ [Since $$x <<0.1$$ it can be neglected]

$$5.88 \times 0.1 \times 10^{-6} =x^2$$

$$x=7.6 \times 10^{-4}$$

%$$ Pyridine =\cfrac {7.6 \times 10^{-4}}{0.1} \times 100 = 0.77\%$$

$$ \because \left (\cfrac {dissociated \ ions}{concentration \ of \ solution }\times 100\right)$$
Hence, the correct option is $$A$$
Question 4
1 / -0

The amount of zinc required to produce $$224$$ ml of $$H_2$$ at NTP on treatment with dilute $$H_2SO_4$$ solution will be:

A
$$0.65$$g

B
$$0.065$$ g

C
$$65$$g

D
$$6.5$$g

Solution

Let us consider the problem.

zinc reacts with sulfuric acid according to the following formula.

$$Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}$$

Therefore, $$1$$ mole of zinc reacts to give $$1$$ mole of hydrogen

Volume of $$1$$ mole of hydrogen at $$STP$$ is $$22400$$ $$ml$$

$$1$$ mole of $$Zn$$ is $$65$$ grams

Therefore $$0.65$$ grams of zinc reacts to give $$224$$ $$ml$$ of hydrogen.

Hence, the correct option is $$A$$
Question 5
1 / -0

What quantity of lime stone on heating will give $$56\ kg$$ of $$CaO$$?

A
$$1000\ kg$$

B
$$56\ kg$$

C
$$44\ kg$$

D
$$100\ kg$$

Solution

$$\displaystyle CaCO_3 \xrightarrow {\Delta } CaO+CO_2 \uparrow$$

Molar masses of $$\displaystyle CaO$$ and $$\displaystyle CaCO_3$$ are 56 g/mol and 100 g/mol respectively.

Mass of $$\displaystyle CaO = 56 \ kg = 56000 \ g$$

$$\displaystyle (\because 1 \ kg = 1000 \ g)$$

Number of moles of $$\displaystyle CaO= \dfrac { 56000 \ g}{ 56 \ g/mol}=1000 \ mol$$

As per balanced chemical equation, 1 mole of $$\displaystyle CaO$$ is obtained by heating 1 mole of $$\displaystyle CaCO_3$$.

Hence, 1000 moles of $$\displaystyle CaO$$ will be obtained by heating 1000 moles of $$\displaystyle CaCO_3$$.

Mass of $$\displaystyle CaCO_3 = 1000 \ mol \times 100 \ g/mol=10^5 \ g$$

Convert unit from g to kg.

Mass of $$\displaystyle CaCO_3 = 10^5 \ g\times \dfrac {1 \ kg}{1000 \ g}=100 \ kg $$

Hence, the correct option is $$D$$
Question 6
1 / -0

What mass of sodium chloride will react with 34.0 g of silver nitrate to produce 17 g of sodium nitrate and 28.70 g of silver chloride if the law of conservation of mass holds good ?

A
12.35 grams

B
11.70 grams

C
9.32 grams

D
None of these

Solution

$$Nacl + AgNO_{3} \rightarrow NaNO_{3}+Agcl$$
By law of conservation of max
total mass on left
By law of conservation of mass
total mass on right
$$ \Rightarrow x+ 3y = 17+28.7 $$
$$ \boxed{x = 11.70 gm}$$
Question 7
1 / -0

If average molecular wt. of air is 29, then assuming $$N_2$$ and $$O_2$$ gases are there which options are correct regarding composition of air:
i) 75% by mass of $$N_2$$
ii) 75% by moles $$N_2$$
iii) 72.41% by mass of $$N_2$$

A
only (i) is correct

B
only (ii) is correct

C
both (ii) and (iii) are correct

D
both (i) and (ii) are correct

Solution

Average wt. air is $$29$$ $$Nitrogen=\cfrac{x}{10}\times 29\\x=75\%$$
Question 8
1 / -0

$$4.6$$ $$cm^3$$ of methyl alcohol is dissolved in $$25.2$$g of water with density $$0.8$$. Calculate $$\%$$ by mass of methyl alcohol.

A
$$14.72$$

B
$$10.67$$

C
$$12.74$$

D
$$13.45$$

Solution

According to he question:
$$\% \left( {\dfrac{w}{W}} \right) = \dfrac{{Weight\,of\,solute}}{{Weight\,of\,solute + weight\,of\,solvent\,\left( {water} \right)\,}} \times 100$$

Substitutes the values,
$$Density = \dfrac{{mass}}{{volume}}$$

$$mass = 0.8 \times 4.6 = 3.68gm.$$

$$\% \,by\,mass = \dfrac{{3.68}}{{3.68 + 25.2}} \times 100$$

$$\% = 12.74\% \,by\,mass$$

Hence, option C is correct.
Question 9
1 / -0

What mass of $$NaCl$$ would contain the same total number of ions as 245 g of $$MgCl_2$$?

A
245 g

B
225 g

C
263 g

D
None of these

Solution

$$ (24+35.5\times2)$$ gm of $$ mgcl_{2}\rightarrow 3$$ No ion
245 gm of $$ mgcl_{2}\rightarrow \dfrac{3}{(24+70)}\times 245\,gm $$
$$(\dfrac{3}{24+70}) \times 245 $$ ion of Nacl $$ \rightarrow \dfrac{(23+35.5)}{2\times Na} \times \dfrac{3}{(24+70)} \times 245\,gm$$
Question 10
1 / -0

The density (in g $$ml^{-1}$$) of a $$3.60$$M sulphuric acid solution having $$29\%$$ $$H_2SO_4$$ [molar mass $$=98$$g $$mol^{-1}$$] by mass, will be:

A
$$1.64$$

B
$$1.88$$

C
$$1.22$$

D
$$1.45$$

Solution

Let us consider the problem:

Mass $$% $$% of $${H_2}S{O_4} = 29 \%$$

Therefore $$100g$$ solution contains $$29g\ {H_2}S{O_4}$$

Let the density of solution $$\left( {in{\text{ }}g/ml} \right)$$ is $$d$$

Molarity of solution $$= \dfrac{{{\text{Moles of}}\ {H_2}S{O_4}}}{{{\text{Volume of solution(ml)}}}} \times 1000$$

$$\left( {M = 3.60} \right)$$

$$\dfrac{{\dfrac{{29}}{{98}}}}{{\dfrac{{100}}{d}}} \times 1000 = 3.60$$

Hence the answer is $$d = 1.22g/ml$$.

Hence, the correct option is $$C$$

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Some Basic Concepts of Chemistry Test - 51

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Some Basic Concepts of Chemistry Test - 51

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Question 1

$$2.5\times 10^{14}$$

$$5\times 10^{14}$$

$$5\times 10^{13}$$

$$5\times 10^{15}$$

Solution

Question 2

$$\dfrac{2\Delta H}{N_A}$$

$$\dfrac{2N_A}{\Delta H}$$

$$\dfrac{\Delta H}{2N_A}$$

$$\dfrac{N_A}{\Delta H}$$

Solution

Question 3

$$0.77\%$$

$$1.6\%$$

$$0.0060\%$$

$$0.013\%$$

Solution

Question 4

$$0.65$$g

$$0.065$$ g

$$65$$g

$$6.5$$g

Solution

Question 5

$$1000\ kg$$

$$56\ kg$$

$$44\ kg$$

$$100\ kg$$

Solution

Question 6

12.35 grams

11.70 grams

9.32 grams

None of these

Solution

Question 7

only (i) is correct

only (ii) is correct

both (ii) and (iii) are correct

both (i) and (ii) are correct

Solution

Question 8

$$14.72$$

$$10.67$$

$$12.74$$

$$13.45$$

Solution

Question 9

245 g

225 g

263 g

None of these

Solution

Question 10

$$1.64$$

$$1.88$$

$$1.22$$

$$1.45$$

Solution

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