Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

A solution of ethanol in water is 10% by volume. If the solution and pure ethanol have densities of 0.9866 g/cc and 0.785 g/cc respectively. The percent by weight is nearly?

A
7.95%

B
17%

C
9.86%

D
16.2%

Solution

Volume of ethanol = 10 ml
Volume of solution = 100 ml
Weight of ethanol = $$ Volume \times density $$
= $$ 10 \times 0.785 $$
= 7.85 g
Weight of solution = $$ 100 \times 0.9866 $$
= 98.66 g
Weight percent of ethanol = $$\frac{7.85}{98.66}$$ $$\times 100 $$
= $$7.95\%$$
Question 2
1 / -0

$$10g$$ of glucose is dissolved in $$150g$$ of water. The mass percantage of glucose is:

A
$$2.50\%$$

B
$$6.67\%$$

C
$$8.75\%$$

D
$$10\%$$

Solution

$$10g$$ is dissolved in $$150g$$ of water
$$\therefore$$ Mass percentage of glucose= $$\cfrac {10}{150}\times 100$$
$$=6.67$$%
Question 3
1 / -0

A fresh $$H_2O_2$$ solution is labeled as $$11.2V$$. Calculate its concentration in wt/vol percent.

A
$$3.03$$

B
$$6.8$$

C
$$1.7$$

D
$$13.6$$

Solution

$$Density = \dfrac{v}{11.2} * M.wt$$
$$d = \dfrac{11.2}{11.2} * 34 = 34$$

$$\dfrac{wt}{v} = \dfrac{34}{11.2} = 3.03%$$
Question 4
1 / -0

In a test tube, there is $$18\ g$$ of glucose $$(C_{6}H_{12}O_{6})0.08$$ mole of glucose is taken out. Glucose left in the test tube is:

A
$$0.10\ g$$

B
$$0.02\ g$$

C
$$0.10\ mole$$

D
$$3.60\ g$$

Solution

Moles of glucose=$$\cfrac {18}{180}=0.1$$ moles
Moles of glucose left=$$0.1-0.08=0.02$$ moles
$$\therefore$$ Mass of glucose left=$$0.02\times 180$$
=$$3.6g$$
Question 5
1 / -0

The mass of $$50\%(w/w)$$ solution of $$HCl$$ required to react with $$100\ g$$ of $$CaCO_{3}$$ would be?

A
$$73\ g$$

B
$$100\ g$$

C
$$146\ g$$

D
$$200\ g$$

Solution

The reaction is:
$$2HCl+CaCO_3\longrightarrow CaCl_2+CO+3H_2$$
$$2$$ molecules of $$HCl$$ give
$$1$$ molecule of $$CaCO_3$$
(Molar mass)$$_{CaCO_3}=100$$ $$g$$
$$100$$ $$g$$ of $$CaCO_3=1$$ $$mole$$
So, Molar mass of $$HCl=35.5+1=36.5$$
$$\therefore 1$$ $$mole$$ of $$HCl=36.5$$ $$g$$
Since, we get $$2$$ $$moles$$ of $$HCl$$ from $$100$$ $$g(1$$ $$mole)$$ of $$CaCO_3$$
Weight of $$HCl=2(36.5)$$ $$g=73$$ $$g$$
Question 6
1 / -0

One mole of a mixture of CO and $${ CO }_{ 2 }$$ requires exactly 20 g of NaOH in solution for complete conversion of all the $${ CO }_{ 2 }$$ into $${ Na }_{ 2 }{ CO }_{ 3 }$$. How much NaOH would it require for conversion into $${ Na }_{ 2 }{ CO }_{ 3 }$$, if the mixture (one mole) is completely oxidized to $${ CO }_{ 2 }$$?

A
60 g

B
80 g

C
40 g

D
20 g

Solution

let $$x$$ and $$y$$ mass of $$CO$$ and $$CO_{2}$$ be there
$$x+y=1$$
$$2NaOH+CO^{2}\rightarrow Na_{2}CO_{3}+H_{2}O$$
$$n$$ of $$NaOH=20/40=0.5$$
$$\therefore 0.5/2$$ mol $$CO_{2}$$ must be there
$$\Rightarrow 0.25$$ mol $$CO_{2}$$
$$\therefore 0.75$$ mol $$CO$$
If $$CO$$ is oxidised to $$CO_{2}$$
$$NaOH$$ required $$=2\times 0.75\times 40$$
$$=60g$$
Question 7
1 / -0

Equivalent wt.of a metal is $$2.5$$ times higher than oxygen. The ratio of weight of metal to the oxide is:

A
$$0.4$$

B
$$1.4$$

C
$$2.4$$

D
$$3.4$$

Solution

Solution:- (A) 0.4
Given that the equivalent weight of a metal is $$2.5$$ times higher than oxygen.
Equivalent weight of oxygen $$= 8$$
$$\therefore$$ Equivalent weight of metal $$=$$ equivalent wt. of oxygen $$+ \; (2.5 \; \times$$ equivalent weight of oxygen) $$= 8 + 2.5 \times 8 = 28$$
As we know that $$\text{eq. wt.} = \cfrac{\text{mol. wt. } \left( {M}_{m} \right)}{\text{valence factor } \left( n \right)}$$
$$\therefore$$ Molecular weight of metal $$\left( {M}_{m} \right) = 28 n$$
Formula of metal oxide will be- $${M}_{2}{O}_{n}$$
Molecular weight of metal oxide $$= 2 \times {M}_{m} + n \times {M}_{O} = 56n + 16 n = 72 n$$
Ratio of weight of metal to its oxide $$= \cfrac{\text{wt. of metal}}{\text{wt. of metal oxide}} = \cfrac{28n}{72n} = 0.389 \approx 0.4$$
Hence the required answer is $$(A) 0.4$$.
Question 8
1 / -0

$$1.5 gm$$ mixture of $${ SiO }_{ 2 }$$ and $${ Fe }_{ 2 }{ O }_{ 3 }$$ on very strong heating leave a residue weighting $$1.46 gm$$. The reaction responsible for loss of weight is
$${ Fe }_{ 2 }{ O }_{ 3 }(s)\longrightarrow { Fe }_{ 3 }{ O }_{ 4 }(s)+{ O }_{ 2 }(g)$$
What is the percentage by mass of $${ Fe }_{ 2 }{ O }_{ 3 }$$ in original sample?

A
$$80\%$$

B
$$20\%$$

C
$$40\%$$

D
$$60\%$$

Solution

$$3 Fe_2 O_3 \rightarrow 2 Fe_3 O_4 + \dfrac{1}{2} O_2 (g)$$
Molecular mass of $$Fe_2 O_3 = 160$$
Molecular mass of $$Fe_3 O_4 = 232$$
Loss in mass per $$480 gm \, Fe_2 O_3 = 16 gm$$
In this ques, loss of mass = $$0.04 gm$$
Mass of $$Fe_2 O_3 = \dfrac{0.04}{16} \times 480$$
$$ = 1.20 gm$$
$$\therefore \%$$ mass of $$Fe_2 O_3 = \dfrac{1.20}{1.50} \times 100 = 80\%$$
Question 9
1 / -0

$$400mg$$ capsule contains $$100mg$$ of ferrous fumarate. The percentage of iron present in the capsule approximately is: (Mol mass of $$Fe({C}_{4}{H}_{2}{O}_{4})=170$$)

A
$$8.2$$%

B
$$25$$%

C
$$16$$%

D
$$50$$%

Solution

Molar mass of ferrous fumarate $$(C_4H_2FeO_4)=169.9$$ $$g$$
Atomic mass of Iron $$(Fe)=55.8$$
$$\therefore 169.9$$ $$g$$ of $$C_4H_2FeO_4\rightarrow 55.8$$ $$g$$ of $$Fe$$
$$\therefore 100$$ $$mg$$ of $$C_4H_2FeO_4$$ contain:
$$=\cfrac{55.8 g}{169.9 g}\times 100$$ $$mg$$
$$=32.8$$ $$mg$$
Amount of $$C_4H_2FeO_4$$ in $$400$$ $$mg$$ capsule $$=100$$ $$mg$$
$$\therefore$$ Amount of iron in $$400$$ $$mg$$ capsule $$=32.8$$ $$mg$$
Percentage of iron in $$400$$ $$mg$$ capusle $$=100\times \cfrac {Amount\quad of\quad Iron\quad in\quad capsule}{Total\quad mass\quad of\quad capsule}$$
$$=100\times \cfrac{32.8 mg}{400 mg}$$
$$=8.2\%$$
Question 10
1 / -0

The density of a $$3 \,M$$ sodium thiosulphate $$(Na_2S_2O_3)$$ solution is $$1.25 \,gm/ml$$. Calculate the percentage by weight of sodium thiosulphate.

A
$$73.92$$

B
$$65.84$$

C
$$17.14$$

D
$$37.92$$

Solution

Percentage by weight of sodium thiosulphate solution $$(Na_2S_2O_3)$$ will be:
$$=\cfrac{wt.\ of\ Na_2S_2O_3}{wt.\ of\ solution}\times 100$$
Also wt. of $$4Na_2SO_3=$$Molarity$$\times $$ molecular wt.$$=3\times 150$$
$$\%$$ by wt.$$=\cfrac{3\times 158}{1.25\times 100}\times 100=37.92$$

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 53

Result

Some Basic Concepts of Chemistry Test - 53

Score

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Rank

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SHARING IS CARING

Question 1

7.95%

17%

9.86%

16.2%

Solution

Question 2

$$2.50\%$$

$$6.67\%$$

$$8.75\%$$

$$10\%$$

Solution

Question 3

$$3.03$$

$$6.8$$

$$1.7$$

$$13.6$$

Solution

Question 4

$$0.10\ g$$

$$0.02\ g$$

$$0.10\ mole$$

$$3.60\ g$$

Solution

Question 5

$$73\ g$$

$$100\ g$$

$$146\ g$$

$$200\ g$$

Solution

Question 6

60 g

80 g

40 g

20 g

Solution

Question 7

$$0.4$$

$$1.4$$

$$2.4$$

$$3.4$$

Solution

Question 8

$$80\%$$

$$20\%$$

$$40\%$$

$$60\%$$

Solution

Question 9

$$8.2$$%

$$25$$%

$$16$$%

$$50$$%

Solution

Question 10

$$73.92$$

$$65.84$$

$$17.14$$

$$37.92$$

Solution

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