Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 49

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Weekly Quiz Competition

Question 1
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0.48 g of a sample of a compound containing boron and oxygen contains 0.192 g of boron and 0.288 g of oxygen. What will be the percentage composition of the compound?

A
$$60\%$$ and $$40\%$$ B and O respectively

B
$$40\%$$ and $$60\%$$ B and O respectively

C
$$30\%$$ and $$70\%$$ B and O respectively

D
$$70\%$$ and $$30\%$$ B and O respectively

Solution

Percent composition of an element in a compound $$=\cfrac{mass\ of\ element}{total\ mass} \times 100$$
% of boron $$=\cfrac{0.192}{0.48} \times 100=40$$%
% of oxygen $$=\cfrac{0.288}{0.48} \times 100=60$$%
Question 2
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A solution is prepared by adding $$5\ g$$ of a solute $$'X'$$ to $$45\ g$$ of solvent $$'Y'$$. What is the mass percent of solute $$'X'$$?

A
$$10\%$$

B
$$11.1\%$$

C
$$90\%$$

D
$$75\%$$

Solution

In a mixture of X and Y,
Mass % of X=$$\dfrac{Mass\ of\ X}{(Mass\ of\ X+Mass\ of\ Y)}\times 100$$
$$\therefore$$ mass % of X=$$\cfrac{5}{5+45} \times 100=10$$ %
Question 3
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The weight of $$AgCl$$ precipitated when a solution containing $$5.85\ g$$ of $$NaCl$$ is added to a solution containing $$3.4\ g$$ of $$AgNO_3$$ is:

A
$$28\ g$$

B
$$9.25\ g$$

C
$$2.870\ g$$

D
$$58\ g$$

Solution

Balanced reactio is:
$$NaCl+AgNO_3 \rightarrow AgCl+NaNO_3$$
molar mass of $$AgNO_3$$=170 g
molar mass of $$NaCl$$=58.5 g
molar mass of $$AgCl$$=143.5 g
from the reaction 58.5 g $$NaCl$$ reacts with 170 g $$AgNO_3$$
thus 5.85 g $$NaCl$$ will reacts with $$=\dfrac{170}{58.5} \times 5.85=17g\ AgNO_3$$
Given $$AgNO_3$$=3.4 g
therefore $$AgNO_3$$ is completely consumed and is the limiting reagent
170 g $$AgNO_3$$ produces 143.5 g $$AgCl$$
3.4 g will produce =$$\dfrac{143.5}{170} \times 3.4=2.87g\ AgCl$$
Hence, option $$C$$ is correct.
Question 4
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How much mass of sodium acetate is required to make 250 mL of 0.575 molar aqueous solution?

A
$$11.79 g$$

B
$$15.38 g$$

C
$$10.81 g$$

D
$$25.35 g$$

Solution

Sodium acetate is $$C_2H_3O_2Na$$
molar mass = 82 g/mol
Given: molarity, M=0.575 M, volume of solution, V=250 mL
number of moles, n=$$\dfrac{mass}{molar\ mass}=\dfrac{m}{82}$$
Using the relation: $$M=\cfrac{n}{V} \times 1000$$
substitute the values:
$$0.575=\dfrac{m}{250\times 82} \times 1000$$
$$m=11.79$$ g
Question 5
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A compound of magnesium contains $$21.9\%$$ magnesium, $$27.8\%$$ phosphorus and $$50.3\%$$ oxygen. What will be the simplest formula of the compound?

A
$$Mg_2P_2O_7$$

B
$$MgPO_3$$

C
$$Mg_2P_2O_2$$

D
$$MgP_2O_4$$

Solution

Explanation:
Atomic mass of $$Mg$$ = $$24.3\ grams$$
Atomic mass of $$P$$ = $$31\ grams$$
Atomic mass of $$O$$ = $$16\ grams$$
Let mass of the given compound be $$100\ grams$$

Moles of $$Mg$$ = $$\dfrac {21.9}{24.3}$$ = $$0.90$$
Moles of $$P$$ = $$\dfrac {27.8}{31}$$ = $$0.90$$
Moles of $$O$$ = $$\dfrac {50.3}{16}$$ = $$3.14$$
Mole ratio $$=0.90:0.90:3.14$$
To simplify this ratio, divide each by $$0.90$$
$$\therefore$$ Mole ratio $$=1:1:3.5=2:2:7$$
We get the formula for the compound is, $$Mg_2P_2O_7$$
Hence, the correct answer is option $$A$$.
Question 6
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What will be the weight of CO having the same number of oxygen atoms as present in 22 g of $$CO_2$$?

A
28 g

B
22 g

C
44 g

D
72 g

Solution

1 mol of $$CO_2=12+32=44\ g$$

Number of oxygen in 1 molecule of $$CO_2=2$$

Therefore, 44 g $$CO_2$$ contains 2 moles of oxygen

22 g $$CO_2$$ contains $$=\dfrac{2}{44} \times 22=1$$ mole oxygen atom

1 mole of $$CO=12+16=28\ g$$

$$\therefore$$ 28 g $$CO$$ contains 1 mol oxygen atom.

The equal number of moles means an equal number of atoms of oxygen.
Hence, the correct option is $$A$$
Question 7
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Which of the following statement about Avogadro's hypothesis is correct?

A
Under similar conditions of temperature and pressure, gases react with each other in simple ratio.

B
Under similar conditions of temperature and pressure, equal volumes of all gases contain same number of molecules

C
At NTP all gases contain same number of molecules

D
Gases always react with gases only at the given temperature and pressure

Solution

Avogadro’s Law states that under same conditions of temperature and pressure, equal volume of all the gases contain equal number of molecules.
Question 8
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In a mixture of gases, the volume content of a gas is $$0.06\%$$ at STP. Calculate the number of molecules of the gas in 1 L of the mixture.

A
$$1.613 \times 10^{23}$$

B
$$6.023 \times 10^{23}$$

C
$$1.61 \times 10^{27}$$

D
$$1.61 \times 10^{19}$$

Solution

Volume content of gas A in a mixture of A and B =0.06%
1 L of mixture contains= $$\dfrac{0.06}{100} \times 1=6 \times 10^{-4}\ L$$ of A
At STP, 1 mol of gas has volume =22.4 L
number of moles in $$6 \times 10^{-4}$$ L gas at STP=$$\dfrac{6 \times 10^{-4}}{22.4}$$
number of molecules=number of moles $$\times 6.023 \times 10^{23}$$
$$=\dfrac{6 \times 10^{-4}}{22.4} \times 6.023 \times 10^{23}$$
$$=1.61 \times 10^{19}$$ molecules
Question 9
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Total number of atoms present in 34 g of $$NH_3$$ is:

A
$$4 \times 10^{23}$$

B
$$4.8 \times 10^{21}$$

C
$$2 \times 10^{23}$$

D
$$48 \times 10^{23}$$

Solution

Mass of 1 mol or ($$6.022 \times 10^{23}$$ molecules) of $$NH_3=14+3\times 1=17\ g$$

Number o f molecules in 34 g $$NH_3=\frac{34}{17} \times 6.022 \times 10^{23}=12.044\times 10^{23}$$

Number of atoms present in 1 molecule of $$NH_3=4$$

Number of atoms in 34 g $$NH_3=4 \times 12.044 \times 10^{23}$$
$$\approx48 \times 10^{23}$$

option D is correct
Question 10
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Major problems faced by energy section are

A
Rising oil prices

B
Rising oil import bill

C
Sick SEBs

D
All of the above

Solution

$$ \text { Major problems faced by energy sector are- } $$
$$ \begin{array}{l} \text { - Coal production remain key to energy mix. } \\ \text {- Fourth largest consumer of oil and petrolium } \\ \text { in the world. } \end{array} $$
$$ \begin{array}{l} \text { -Relies on imports to meet growing demand of gas } \\ \text { -electricity shortages hurt industrial output. } \\ \text {- Rising of prices, Rising oil import bill, } \\ \text { sick } S E Bs \text { . } \end{array} $$

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Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 49

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Some Basic Concepts of Chemistry Test - 49

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Question 1

$$60\%$$ and $$40\%$$ B and O respectively

$$40\%$$ and $$60\%$$ B and O respectively

$$30\%$$ and $$70\%$$ B and O respectively

$$70\%$$ and $$30\%$$ B and O respectively

Solution

Question 2

$$10\%$$

$$11.1\%$$

$$90\%$$

$$75\%$$

Solution

Question 3

$$28\ g$$

$$9.25\ g$$

$$2.870\ g$$

$$58\ g$$

Solution

Question 4

$$11.79 g$$

$$15.38 g$$

$$10.81 g$$

$$25.35 g$$

Solution

Question 5

$$Mg_2P_2O_7$$

$$MgPO_3$$

$$Mg_2P_2O_2$$

$$MgP_2O_4$$

Solution

Question 6

28 g

22 g

44 g

72 g

Solution

Question 7

Under similar conditions of temperature and pressure, gases react with each other in simple ratio.

Under similar conditions of temperature and pressure, equal volumes of all gases contain same number of molecules

At NTP all gases contain same number of molecules

Gases always react with gases only at the given temperature and pressure

Solution

Question 8

$$1.613 \times 10^{23}$$

$$6.023 \times 10^{23}$$

$$1.61 \times 10^{27}$$

$$1.61 \times 10^{19}$$

Solution

Question 9

$$4 \times 10^{23}$$

$$4.8 \times 10^{21}$$

$$2 \times 10^{23}$$

$$48 \times 10^{23}$$

Solution

Question 10

Rising oil prices

Rising oil import bill

Sick SEBs

All of the above

Solution

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