Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 47

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Question 1
1 / -0

A mixture contains equimolar quantities of carbonates of two bivalent metals. One metal is present to the extent of $$13.5\%$$ by weight in the mixture and $$2.58$$g of the mixture on heating leaves a residue of $$1.35$$g. What percentage by weight of the other metal is there?

A
$$31.6$$

B
$$21.3$$

C
$$19.2$$

D
$$17.8$$

Solution

Let equimolar metal carbonates have mole be $$n$$
Let Metal be $$M_1$$ and $$M_2$$
As $${ M }_{ 1 }{ CO }_{ 3 }\underrightarrow { \Delta } { M }_{ 1 }O+{ CO }_{ { 2 }_{ (g) } }\uparrow $$
$${ M }_{ 2 }{ CO }_{ 3 }\underrightarrow { \Delta } { M }_{ 2 }O+{ CO }_{ { 2 }_{ (g) } }\uparrow $$
Let molecular wt. of $${ M }_{ 1 }\rightarrow { w }_{ 1 }$$
molecular wt. of $${ M }_{ 2 }\rightarrow { w }_{ 2 }$$
for % composition of metal $${ M }_{ 1 }=\cfrac { n{ w }_{ 1 } }{ n\left( { w }_{ 1 }+60 \right) +n\left( { w }_{ 2 }+60 \right) } =0.135\quad -(1)$$
$$\Rightarrow$$ we have $$n\left( { w }_{ 1 }+60 \right) +n\left( { w }_{ 2 }+60 \right) =2.58gm\quad -(2)$$
$$CO_2$$ librated$$=2.58-1.35=1.23gm$$
$$\therefore$$ wt. of $$CO_2$$ in mixture$$=n(44)+n(44)$$
$$=88n\quad gm$$
$$88n\quad gm=1.23gm$$
$$n=\cfrac {1.23}{88} =0.01397$$
Put $$n=0.01397$$ mol in $$(1)$$ we get
$$\cfrac {0.01397 w_1}{2.58} =0.135$$
$$[w_1 =24.93gm]$$
From $$(2)$$
$$n(w_2 +w_1 +120)=2.58$$
$$0.01397(24.93+ w_2 +120)=2.58$$
$$2.0246+0.01397 w_2 =2.58$$
$$[w_2 =39.75gm]$$
% composition of $$M_2 =\cfrac { n{ w }_{ 2 } }{ 2.58 } =\cfrac { 0.01397\times 39.75 }{ 2.58 } =0.215$$
$$=21.5$$%
Question 2
1 / -0

A precipitate of AgCl and AgBr has the mass 0.4066 g. On heating a current of chlorine, the AgBr is converted to AgCl and the mixture loses 0.0725 in mass. The % of mass in the original mixture is (as the nearest integer) :

A
2 %

B
6 %

C
5%

D
95%

Solution

Option B is correct
Question 3
1 / -0

The leaves of a green bean plant were exposed to carbon dioxide containing the radioactive isotope of carbon $$^{14}C$$ for few hours in sunlight. A section of the plant's stem was then immediately obtained and placed on an X-ray film that blackens where exposed to radioactivity. The following diagrams show the cross-section of the stem, and the resulting autoradiograph. Why is structure $$P$$ shown radioactive?

A
$$P$$ transported radioactive carbohydrates from leaves to other plant parts

B
Radioactive carbon dioxide gets dissolved in cell sap and is transported along with water and minerals by structure $$P$$

C
The radioactive carbon is incorporated only in the tissues constituting structure $$P$$

D
None of these
Question 4
1 / -0

Decreasing order of atomic weight is correct of the elements given below?

A
$$Fe > Co > Ni$$

B
$$Ni > Co > Fe$$

C
$$Co > Ni > Fe$$

D
$$Co > Fe > Ni$$

Solution

Atomic weight of Iron $$\approx 56$$ u
Atomic weight of Cobalt $$\approx 59$$ u
Atomic weight of Nickel $$\approx 58.7$$ u
So, the order of atomic weights is $$Co>Ni>Fe$$.
Option $$C$$ is correct.
Question 5
1 / -0

The molecular weights of $$O_2$$ and $$N_2$$ are $$32$$ and $$28$$ respectively. At $$15^0$$C, the pressure of $$1$$gm $$O_2$$ will be the same as that of $$1$$ gm $$N_2$$ in the same bottle at the temperature:

A
$$-21^0$$C

B
$$-13^0$$C

C
$$15^0$$C

D
$$56.4^0$$V

Solution

Temperature of $$O_{2}$$ $$=273+15=298 K$$

Temperature of $$N_{2}$$ $$=x$$

As $$n$$ $$\propto$$ $$\dfrac 1T$$, as pressure is constant.

$$\Longrightarrow { n }_{ 1 }{ T }_{ 1 }={ n }_{ 2 }{ T }_{ 2 }\Longrightarrow \left( \cfrac { 1 }{ 32 } \right) \left( 298 \right) =\left( \cfrac { 1 }{ 28 } \right) \left( x \right) \\ \Longrightarrow x=\cfrac { 28 }{ 32 } \times 298=260.75K$$

Temperature in $$^oC=260.75-273=-13 ^oC$$

Hence, option $$B$$ is correct.
Question 6
1 / -0

Calculate the total number of oxygen atoms present in $$1.80g$$ of glucose $$({C}_{6}{H}_{12}{O}_{6})$$.

A
$$46.138\times10^{21}$$

B
$$36.138\times10^{21}$$

C
$$26.138\times10^{21}$$

D
$$6.138\times10^{21}$$

Solution

$$Glucose\longrightarrow C_6H_{12}O_6$$
$$Number\ of\ moles=\dfrac{mass}{molecular\ mass}$$
$$=\dfrac{1.8}{180}$$
$$=0.01\ mol$$
Now, there are 6 atoms of oxygen
Total number of oxygen atoms$$=$$number of atoms of oxygen$$\times$$number of moles$$\times$$Avagadro's constant
$$=6\times0.01\times6.023\times10^{23}$$
$$=36.138\times10^{21}$$
Question 7
1 / -0

What will be the mass percentage of a resulting solution prepared by mixing $$15\%$$ (w/w) 500 g aqueous solution of the area with $$25\%$$ (w/w) 400g aqueous solution of it?

A
$$18\%$$

B
$$20\%$$

C
$$25\%$$

D
$$15\%$$

Solution

Given $$({w/w})\%$$ of $$1^{st}$$ solution $$=15$$
$$\implies 15=\cfrac{500}{Weight\quad of\quad solution(x_1)}\times 100$$
$$\implies x_1=\cfrac{(500)(100)}{15}$$
$$\implies x_1=\cfrac{10^{4}}{3}$$
Given $$({w/w})\%$$ of $$2^{nd}$$ solution $$=25$$
$$\implies 25=\cfrac{400}{Weight\quad of\quad solution(x_2)}\times 100$$
$$\implies x_2=\cfrac{(400)(100)}{25}$$
$$\implies x_2=1600$$
Total $$({w/w})\%=\left[\cfrac{500+400}{\cfrac{10^4}{3}+1600}\right]\times 100=\cfrac{900}{33.3+16}=\cfrac{900}{49.3}\simeq 18$$
$$\therefore$$ Weight by weight percentage of resulting solution is $$18\%$$
Question 8
1 / -0

The critical volume of a gas is 0.036 $$lit. mol^{-1}$$. The radius of the molecule will be (in cm):
(Avogadro Number = $$6 \times 10^{23}$$)

A
$$\displaystyle \left( \frac{9}{4 \pi} \times 10^{-23} \right)^{\dfrac{1}{3}}$$

B
$$\displaystyle \left( \frac{8 \pi}{3} \times 10^{-23} \right)^{\dfrac{1}{3}}$$

C
$$\displaystyle \left( \frac{3}{8 \pi} \times 10^{-23} \right)^{\dfrac{1}{3}}$$

D
none of these

Solution

Critical volume of gas=3b [for 1 molecule]
For 1 mole, critical volume=0.036 liters
$$\Longrightarrow 0.036\times { 10 }^{ 3 }{ cm }^{ 3 }=3\left( \cfrac { 4 }{ 3 } \pi { r }^{ 3 } \right) \times { N }_{ A } \\ \Longrightarrow r={ \left( \cfrac { 36 }{ 24\pi } \times 10^{ -23 } \right) }^{ \cfrac { 1 }{ 3 } }={ \left( \cfrac { 3 }{ 8\pi } \times 10^{ -23 } \right) }^{ \cfrac { 1 }{ 3 } }$$
Question 9
1 / -0

What do carbon credits signify?

A
Credits given in the course of carbon products sales

B
Entitlements to emit certain quantity of green house gases

C
Permissible amount of Carbon dioxide in the atmosphere

D
The extent of carbon required to ensure sustainable development

Solution

A carbon credit is a generic term for any tradable certificate or permit representing the right to emit one tonne of carbon dioxide or the mass of another greenhouse gas with a carbon dioxide equivalent, equivalent to one tonne of carbon dioxide.
Question 10
1 / -0

What mass of propene is formed from 34.0 g of iodopropane on heating with ethanolic $$KOH$$, if the yield is 36 %?

A
17.2 g

B
8.4 9

C
3.024 g

D
1.72 g

Solution

$$CH_3CH_2CH_2I+KOH \longrightarrow CH_3-CH=CH_2+KI+H_2O$$
Moles of Iodopropane $$=\cfrac{34}{170}=0.2$$ $$moles$$
$$1$$ $$mole$$ of iodopropane gives $$1$$ $$mole$$ propene but yield is $$36\%$$.
So, $$1$$ $$mole$$ of iodopropane will give $$0.36$$ $$mole$$ of propene.
$$0.2$$ $$mole$$ of iodopropane will give $$0.2\times 0.36=0.072$$ $$mole$$ of propene.
Mass of propene $$=42\times 0.072=3.024$$ $$gram$$

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Some Basic Concepts of Chemistry Test - 47

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Some Basic Concepts of Chemistry Test - 47

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Question 1

$$31.6$$

$$21.3$$

$$19.2$$

$$17.8$$

Solution

Question 2

2 %

6 %

5%

95%

Solution

Question 3

$$P$$ transported radioactive carbohydrates from leaves to other plant parts

Radioactive carbon dioxide gets dissolved in cell sap and is transported along with water and minerals by structure $$P$$

The radioactive carbon is incorporated only in the tissues constituting structure $$P$$

None of these

Question 4

$$Fe > Co > Ni$$

$$Ni > Co > Fe$$

$$Co > Ni > Fe$$

$$Co > Fe > Ni$$

Solution

Question 5

$$-21^0$$C

$$-13^0$$C

$$15^0$$C

$$56.4^0$$V

Solution

Question 6

$$46.138\times10^{21}$$

$$36.138\times10^{21}$$

$$26.138\times10^{21}$$

$$6.138\times10^{21}$$

Solution

Question 7

$$18\%$$

$$20\%$$

$$25\%$$

$$15\%$$

Solution

Question 8

$$\displaystyle \left( \frac{9}{4 \pi} \times 10^{-23} \right)^{\dfrac{1}{3}}$$

$$\displaystyle \left( \frac{8 \pi}{3} \times 10^{-23} \right)^{\dfrac{1}{3}}$$

$$\displaystyle \left( \frac{3}{8 \pi} \times 10^{-23} \right)^{\dfrac{1}{3}}$$

none of these

Solution

Question 9

Credits given in the course of carbon products sales

Entitlements to emit certain quantity of green house gases

Permissible amount of Carbon dioxide in the atmosphere

The extent of carbon required to ensure sustainable development

Solution

Question 10

17.2 g

8.4 9

3.024 g

1.72 g

Solution

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