Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 74

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Question 1
1 / -0

Flask I contains a gaseous mixture of $$He(g)$$, $$H_2(g)$$, in the molar ratio of $$1:\sqrt{2}$$. The gas mixture is allowed to effuse through a fine orifice in the flask l and is collected in flask ll. What is the average molar mass of gas mixture initially collected in flask ll?

A
3.20

B
2.33

C
4.00

D
None of these

Solution

$$(A)$$ Given
$$\dfrac{\ ^{n}H_{e}}{\ ^{n}H_{2}}=\dfrac{1}{\sqrt{2}}\Rightarrow \ ^{n}H_{e}=\dfrac{\ ^{n}H_{2}}{\sqrt{2}}$$
Let $$\ ^{n}H_{2}=x$$ then $$\ ^{n}H_{e}=\dfrac{x}{\sqrt{2}}$$
mole fraction of $$H_{2}=\dfrac{x}{\dfrac{x}{\sqrt{2}}+x}=\dfrac{\sqrt{2}}{1+\sqrt{2}}=Y_{A}$$
mole fraction of $$H_{e}=\dfrac{1}{1+\sqrt{2}}=Y_{B}$$
arg molecular weight $$(\bar{M}w)=M_{H_{e}}Y_{B}+M_{H_{2}}Y_{A}$$
Question 2
1 / -0

32g of a sample of $${ FeSO }_{ 4 }.{ 7H }_{ 2 }O$$ where the dissolved in dilute Sulphuric acid and water and it's volume was made up to 1 litre, 25 mL of this solution required 20 mL of 0.02 $$M\ KMn{ O }_{ 4 }$$ solution of complete oxidation. Calculate the weight % of $$FeS{ O }_{ 4 }.{ 7H }_{ 2 }O$$ in the sample.

A
34.75

B
69.5

C
89.5

D
None of these

Solution

$$M_{eq}$$ of $$FeSO_4.7H_2O$$ in $$25mL$$=$$M_{eq}$$ of $$KMnO_4$$
$$=20\times 0.2 \times 5=80$$
So,$$M_{eq}$$ of $$FeSO_4.7H_2O$$ in 1000mL$$=20\times 0.2 \times 5\times\dfrac{1000}{25}=80$$
$$\implies \dfrac{W}{278}\times 1\times 1000=80$$
$$\implies W=22.24$$
So,%$$feSO_4.7H_2O=\dfrac{22.4}{32}\times 100=69.5$$
Question 3
1 / -0

Natural water contains about $$0.02\%$$ $$D_2O$$ (heavy water). When it is enriched to $$20\%$$ (by volume), calculate the fraction of weight due to neutrons in 1 mole of sample.

A
$$0.355$$

B
$$0.444$$

C
$$0.455$$

D
$$0.500$$

Solution

Assume total sample is of $$100gm$$
Now, $$20gm = D_2O$$ and $$80gm = H_2O$$,
Mass of neutrons in $$80gm$$ of $$H_2O = \dfrac{320}{9}$$
mass of neutrons in $$20gm$$ of $$D_2O = \dfrac{100}{11}$$
Fraction of weight $$= \dfrac{\dfrac{320}{11}+\dfrac{100}{11}}{100} = .455$$
Question 4
1 / -0

Find out the weight of solute (M. wt. 60) that is required to dissolve in $$180g$$ water to reduce the vapour pressure to $$\dfrac{4}{5}th$$ of pure water:

A
$$120g$$

B
$$150g$$

C
$$300g$$

D
$$75g$$

Solution

Relative lowering in vapour pressure$$=\cfrac{P_{0}-P_{S}}{P_{0}}=\cfrac{W_{1}\times M_{2}}{M_{1}\times W_{2}}$$
where 1 & 2 denote solute and solvent respectively.
Given, $$P_{S}=\cfrac{4}{5}P_{0}$$
$$\Rightarrow \cfrac{1}{5}=\cfrac{W\times 18}{60\times 180}$$
$$\therefore W=120\,g$$
Question 5
1 / -0

What mass of $$AgI$$ will dissolve in $$1.0L$$ of $$1.0M$$ $${NH}_{3}$$? neglect change in conc. of $${NH}_{3}$$
(Given : $${K}_{sp}=1.5\times {10}^{-16};{K}_{f}[Ag{({NH}_{3})}_{2}^{+}]$$; (At.mass $$Ag=108,I=127$$)

A
$$4.9\times {10}^{-5}g$$

B
$$0.0056g$$

C
$$0.035g$$

D
$$0.011g$$

Solution
Question 6
1 / -0

4.24 g of an organic compound on combustion produces 8.45 g of $${ CO }_{ 2 }$$ and 3.46 g of water. The mass percentage of $$C$$ and $$H$$ in the compound respectively are:

A
$$ 27.2, 18.2$$

B
$$54.4,9.1$$

C
$$9.1,54.4$$

D
$$18.2,27.2$$

Solution

% of carbon $$=\dfrac { 12 }{ 44 } \times \dfrac { Weight\ of\ { CO }_{ 2 } }{ Weight\ of\ organic\ compound } \times 100$$

$$=\dfrac { 12 }{ 44 } \times \dfrac { 3.45 }{ 4.24 } \times 100=54.35\approx 54.4$$

% of Hydrogen $$=\dfrac { 2 }{ 18 } \times \dfrac { weight\quad of\quad { H }_{ 2 }O }{ weight\quad of\quad organic\quad compound } \times 100$$

$$=\dfrac { 2 }{ 18 } \times \dfrac { 3.46 }{ 4.24 } \times 100=9.06=9.1$$

Ans :- $$C=54.4$$%, $$H=9.1$$%
Hence, option $$B$$ is correct.
Question 7
1 / -0

If a solution contains 34.2 g of sugar in 100 g of water. Find the concentration inn terms of mass by mass percentage of the solution?

A
20 %

B
25.48 %

C
50 %

D
10 %

Solution

Mass of sugar$$=34.2\,gm$$

Mass of the solution$$34.2+100=134.2\,gm$$

$$\therefore $$ Concentration $$=\cfrac{34.2}{134.2}\times 100=25.48\%$$
Question 8
1 / -0

If molecular weight of glucose-1-phosphate is 260 and its density is 1.5 gm $$ cm^{-3} $$. What is the average volume occupied by a molecule of this compound?

A
$$ 4.3 \times 10^{-23}cm^{3} $$

B
$$0.67\ cm^{3} $$

C
$$ 0.17 \times 10^{-23}cm^{3} $$

D
$$ 29 \times 10^{-23}cm^{3} $$

Solution

We get volume of $$1$$ mole $$=\dfrac { mass }{ density } =\dfrac { 260 }{ 1.5 } =173.333\dfrac { gm }{ { cm }^{ 3 } } $$
We know $$1$$ mole $$=6.022\times { 10 }^{ 23 }$$ molecules $$=$$ Avagadra number
$$\dfrac { 173.333 }{ 6.022\times { 10 }^{ 23 } } \approx 29\times { 10 }^{ -23 }{ cm }^{ 3 }$$
Ans :- Option D.
Question 9
1 / -0

What percent of a sample of nitrogen must be allowed to escape if its temperature, pressure, and volume are to be changed from $$220^oC$$, $$3$$atm and $$1.65$$litre to $$110^oC$$, $$0.7$$ atm and $$1.00$$ litre respectively?

A
$$81.8$$%

B
$$71.8$$%

C
$$76.8$$%

D
$$86.8$$%

Solution

Let $$n_{i}$$ and $$n_{f}$$ be initial & final moles of nitrogen respectively.

$$\because PV=nRT$$

$$\therefore 3\times 1.65=n_{i}\times R\times 493$$

$$n_{i}=\cfrac{4.95}{493 R}$$

Similarly, $$0.7\times 1=n_{f}\times R\times 383$$

$$\therefore n_{f}=\cfrac{0.7}{383\,R}$$

$$\therefore \%$$ nitrogen escaped$$=\cfrac{(n_{i}-n_{f})}{n_{i}}\times 100$$

$$=\cfrac{\left (\cfrac{4.95}{493R}-\cfrac{0.7}{383R} \right)}{\cfrac{4.95}{493R}}\times 100$$

$$=81.8\%$$ $$\because R= 0.082057 L.atm.K^{-1}.mol^{-1}$$

Hence, the correct option is $$A$$
Question 10
1 / -0

When 15gm of a mixture of $$NaCl$$ and $$Na_2CO_3$$ is heated with dilute hydrochloric acid, 2.5gm of $$CO_2$$ is evolved at NTP. Calculate percentage composition of $$Na_2CO_3$$ in a mixture.

A
$$59.85\%$$

B
$$40.15\%$$

C
$$30\%$$

D
$$70\%$$

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 74

Result

Some Basic Concepts of Chemistry Test - 74

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SHARING IS CARING

Question 1

3.20

2.33

4.00

None of these

Solution

Question 2

34.75

69.5

89.5

None of these

Solution

Question 3

$$0.355$$

$$0.444$$

$$0.455$$

$$0.500$$

Solution

Question 4

$$120g$$

$$150g$$

$$300g$$

$$75g$$

Solution

Question 5

$$4.9\times {10}^{-5}g$$

$$0.0056g$$

$$0.035g$$

$$0.011g$$

Solution

Question 6

$$ 27.2, 18.2$$

$$54.4,9.1$$

$$9.1,54.4$$

$$18.2,27.2$$

Solution

Question 7

20 %

25.48 %

50 %

10 %

Solution

Question 8

$$ 4.3 \times 10^{-23}cm^{3} $$

$$0.67\ cm^{3} $$

$$ 0.17 \times 10^{-23}cm^{3} $$

$$ 29 \times 10^{-23}cm^{3} $$

Solution

Question 9

$$81.8$$%

$$71.8$$%

$$76.8$$%

$$86.8$$%

Solution

Question 10

$$59.85\%$$

$$40.15\%$$

$$30\%$$

$$70\%$$

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