Some Basic Concepts of Chemistry Test

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Some Basic Concepts of Chemistry Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

A balloon contains a certain mass of neon gas. The temperature is kept constant, and the same mass of argon gas is added to the balloon. What happens?

A
The balloon doubles in volume.

B
The volume of the balloon expands by more than two times.

C
The volume of the balloon expands by less than two times.

D
The balloon stays the same size but the pressure increases.
Question 2
1 / -0

A $250$ g sample of a hydrated salt was heated at ${110}^{o}C$ until all water was driven off. The remaining solid weighed $160$ g. From these data, the percent of water by weight in the original sample can be correctly calculated as :

A
$\cfrac { 160 }{ 340 } \times 100$ %

B
$\cfrac { 90 }{ 340 } \times 100$ %

C
$\cfrac { 160 }{ 250 } \times 100$ %

D
$\cfrac { 90 }{ 250 } \times 100$ %

E
$\cfrac { 90 }{ 160 } \times 100$ %

Solution

The percent of water by weight in the original sample can be correctly calculated as :
$= \cfrac { \text { mass of water } }{ \text { mass of hydrated sample } } \times 100$
$= \cfrac { \text { mass of hydrated sample } - \text { mass of anhydrous sample } }{ \text { mass of hydrated sample } } \times 100$
$= \cfrac { 250 -160 }{ 250 } \times 100$
$= \cfrac { 90 }{ 250 } \times 100$
Question 3
1 / -0

What is the volume occupied by 17.75 g of $Cl_2$ at STP?

A
$5.6l$

B
$22.4l$

C
$11.2l$

D
$2.8l$

Solution

In order to solve this problem, we would use the ideal gas law formula PV = nRT
at STP standard temp and pressure is 1 atm and 273k
we have
P = 1atm, n = $\dfrac{17.75}{70.90}$ mole = 0.250, R = $0.0821$ $atm*L/mol*K$ , T= $273K$
V = $\dfrac {nRT}{P}$
= $0.250\times 0.0821\times 273$
= $5.6L$
answer is A
Question 4
1 / -0

What is the percent composition by mass of aluminium in a compound of aluminium and oxygen if the mole ratio of $Al : O$ is $2:3$ ?

A
$37$ % $Al$

B
$47$ % $Al$

C
$53$ % $Al$

D
$63$ % $Al$

E
$74$ % $Al$

Solution

In a compound of aluminium and oxygen the mole ratio of $Al : O$ is $2:3$
The atomic masses of $Al$ and $O$ are 27.0 g/mol and 16 g/mol respectively.
2 moles of Al $= 2 \times 27.0 = 54.0$ g
3 moles of O $=3 \times 48.0$
Hence, the percent composition by mass of aluminium is $\dfrac {54.0}{54.0+48.0} \times 100= 53$ %.
Question 5
1 / -0

Two gases A and B are taken in same volume containers under similar conditions of temperature and pressure. In container A, there are '2N' molecules of gas A. How many number molecules does container B have?

A
2N

B
4N

C
N

D
8N

Solution

Avogadro's law: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
Thus ans is 2N.
Question 6
1 / -0

What is the molar mass of ammonium carbonate ${({NH}_{4})}_{2}{CO}_{3}$ ?

A
$48\ g/mol$

B
$96\ g/mol$

C
$82\ g/mol$

D
$78\ g/mol$

E
$192\ g/mol$

Solution

$(NH_4)_2CO_3$ is the chemical formula of ammonium carbonate.

$N = 14 \times (2) = 28$
$H = 1 \times (4 \times 2) = 8$
$C = 12 \times 1 = 12$
$O = 16 \times 3 = 48$

Molar mass $= 28 + 8 + 12 + 48 = 96\ g/mol$
Question 7
1 / -0

How many grams are there in a $7.0$ mole sample of sodium hydroxide?

A
$40.0g$

B
$140.0\ g$

C
$280.0\ g$

D
$340.0\ g$

E
$420.0\ g$

Solution

The formula of sodium hydroxide is $NaOH$
Atomic weight of $Na$ = $23\ {g}/{mol}$
Atomic weight of $O$ = $16\ {g}/{mol}$
Atomic weight of $H$ = $1\ {g}/{mol}$
$\therefore$ molecular weight of $NaOH$ = atomic mass of $Na$ + atomic mass of $O$ + atomic mass of $H$
= $23 + 16 + 1 = 40\ {g}/{mol}$
Therefore, 1 mole of $NaOH = 40\ g/mol$
7 moles of $NaOH = 7 \ mole \times 40\ g/mol= 280g$

Hence, option $B$ is correct.
Question 8
1 / -0

22.4 litres of a gas at STP weighs 16 g. Identify the gas.

A
Methane

B
Carbon monoxide

C
Ethane

D
Oxygen

Solution

At STP, 1 mole of gas occupies $22.4l$ .
Given gas occupies $22.4l$ at STP and has weight 16 g.
So molecular weight of gas is 16.
Thus Ans: Methane (Mol. weight of methane is 16)
Question 9
1 / -0

Hydrogen, oxygen and carbon dioxide are taken in containers of $2 l$ volume each. Compare the ratio of the number of molecules of the three gases respectively, under same conditions of temperature and pressure.

A
1:8:22

B
1:1:1

C
1:16:44

D
1:8:44

Solution

Using ideal gas law: PV =nRT
n = $\dfrac{V}{RT}$
here P, R and T= Constant, V = 2l
hence number of moles also equal
the ration of moles equal to 1:1:1
answer is B
Question 10
1 / -0

A mixture contains $5.0\ g$ of compound $X$ and $20.0\ g$ of compound $Y$ .What is the percent by mass of compound $X$ ?

A
$75$

B
$25$

C
$80$

D
$20$

Solution

Given,
Mass of X $= 5g = m_x$
Mass of Y $= 20g = m_y$
$\%$ Mass of X is defined as ratio of mass of X and total mass of mixture.
$\%$ Mass of X $= (\dfrac{m_x}{m_x + m_y}) \times 100 = (\dfrac{5}{5 + 20}) \times 100 = 20\%$

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Some Basic Concepts of Chemistry Test - 40

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Some Basic Concepts of Chemistry Test - 40

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Question 1

The balloon doubles in volume.

The volume of the balloon expands by more than two times.

The volume of the balloon expands by less than two times.

The balloon stays the same size but the pressure increases.

Question 2

160340×100\cfrac { 160 }{ 340 } \times 100340160​×100 %

90340×100\cfrac { 90 }{ 340 } \times 10034090​×100 %

160250×100\cfrac { 160 }{ 250 } \times 100250160​×100 %

90250×100\cfrac { 90 }{ 250 } \times 10025090​×100 %

90160×100\cfrac { 90 }{ 160 } \times 10016090​×100 %

Solution

Question 3

5.6l5.6l5.6l

22.4l22.4l22.4l

11.2l11.2l11.2l

2.8l2.8l2.8l

Solution

Question 4

373737% AlAlAl

474747% AlAlAl

535353% AlAlAl

636363% AlAlAl

747474% AlAlAl

Solution

Question 5

2N

4N

N

8N

Solution

Question 6

48 g/mol48\ g/mol48 g/mol

96 g/mol96\ g/mol96 g/mol

82 g/mol82\ g/mol82 g/mol

78 g/mol78\ g/mol78 g/mol

192 g/mol192\ g/mol192 g/mol

Solution

Question 7

40.0g40.0g40.0g

140.0 g140.0\ g140.0 g

280.0 g280.0\ g280.0 g

340.0 g340.0\ g340.0 g

420.0 g420.0\ g420.0 g

Solution

Question 8

Methane

Carbon monoxide

Ethane

Oxygen

Solution

Question 9

1:8:22

1:1:1

1:16:44

1:8:44

Solution

Question 10

757575

252525

808080

202020

Solution

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$\cfrac { 160 }{ 340 } \times 100$ %

$\cfrac { 90 }{ 340 } \times 100$ %

$\cfrac { 160 }{ 250 } \times 100$ %

$\cfrac { 90 }{ 250 } \times 100$ %

$\cfrac { 90 }{ 160 } \times 100$ %

$5.6l$

$22.4l$

$11.2l$

$2.8l$

$37$ % $Al$

$47$ % $Al$

$53$ % $Al$

$63$ % $Al$

$74$ % $Al$

$48\ g/mol$

$96\ g/mol$

$82\ g/mol$

$78\ g/mol$

$192\ g/mol$

$40.0g$

$140.0\ g$

$280.0\ g$

$340.0\ g$

$420.0\ g$

$75$

$25$

$80$

$20$