Some Basic Concepts of Chemistry Test

Result

Some Basic Concepts of Chemistry Test - 52

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The mass of a molecule of the compound $$C_{60}H_{122}$$ is __________.

A
$$1.4\times 10^{-21}$$ g

B
$$1.09\times 10^{-21}$$ g

C
$$5.025\times 10^{23}$$ g

D
$$16.023\times 10^{23}$$ g

Solution

Molecular mass of $$C_{60}H_{122}=(60 \times 12+1 \times 122)=720+122=842$$

Hence, one mole contains $$6.022 \times 10^{23}$$ molecules

Therefore,

Mass of one molecule $$=\frac{842}{6.022 \times 10^{23}}$$

$$=1.4 \times 10^{-21}\;g$$

The correct option is A.
Question 2
1 / -0

According to the $$CaCO_3+HCl\rightarrow CaCl_2 +CO_2 + H_2O$$. What mass of $$CaCO_3$$ is required to react completely with $$25$$ml. of $$0.75M$$ $$HCl$$?

A
0.9945 g

B
0.6645 g

C
0.9375 g

D
0.8555 g

Solution

Using morality, Number of moles are calculated for $$1000\;ml$$ or $$1\;L$$ of the solution.
$$1000\;ml$$ contains $$0.75\;M \;HCl$$
$$HCl=0.75$$ mole.
Therefore, $$25\;ml$$ of $$0.75\;M$$ HCl will contains $$HCl =\dfrac{0.75 \times 25}{1000}=0.01875\;mole$$
$$2$$ moles of $$HCl$$ reacts with $$=1\;mole$$ of $$CaCO_3$$
Therefore,
$$0.01875$$ mole of $$HCl$$ will react with $$=\dfrac{1 \times 0.01875}{2}$$
$$=0.009375\;mole$$
Molar mass of $$CaCO_3=100\;g$$
Hence, the mass of $$0.009375$$ of $$CaCO_3=no.\;of\;moles \times molar\;mass=0.9375\;g$$
Question 3
1 / -0

Two flasks $$A$$ and $$B$$ of $$500$$ mL each are respectively filled with $$O_2$$ and $$SO_2$$ at $$300$$K and $$1$$ atm pressure. The flasks contain:

A
the same number of atoms

B
the same number of molecules

C
more number of moles in flask $$A$$ as compared to flask $$B$$

D
the same mass of gases

Solution

According to Avogadro's Hypothesis: Equal volumes of all gases contain equal number of atoms under similar conditions of temperature and pressure. So, the correct option will be '$$A$$'.
Question 4
1 / -0

15 g of methyl alcohol is present in 100 mL of solution. If the density of solution is 0.96 g $${ mL }^{ -1 }$$, Calculate the mass percentage of methyl alcohol in solution.

A
45.23%

B
15.66%

C
25.36%

D
None of these

Solution

Mass of methyl alcohol = 15 g
Volume of the solution = 100 ml
Density = $$0.96 gm L^-$$$$^1$$
So. Density = $$\dfrac{mass}{volume}$$
mass = 100 $$\times$$ 0.90 = 96 g
Mass percentage of methyl alcohol = $$\dfrac{ mass\ of\ methyl\ alcohol}{mass\ of\ solution}$$ $$\times$$ 100
= $$\dfrac{1500}{96}=15.66\%$$
Question 5
1 / -0

$$AgNO_3 $$ sample is 85% by mass. To prepare $$12.5\ m $$ of 0.05 molar $$AgNO_3$$ solution, $$AgNO_3$$ sample required is:

A
2.36 g

B
1.25 g

C
1.49 g

D
none of these

Solution

Given, percentage purity of $$\mathrm{AgNO}_{3}$$ Sample $$=85 \%$$
We are to prepare $$125 \mathrm{~mL}$$ solution haring molarity $$=0.05$$.

Thus, moles of pure $$\mathrm{AgNO}_{3}$$ needed = Volume $$\times$$ molarity

$$\begin{array}{l}=\dfrac{125}{1000} \times 0.05\mathrm{mo}\\=6.25\times 10^{-3} \mathrm{~mol} .\end{array}$$

weight of pure $$\operatorname{AgN} O_{3}$$ needed $$=\left(6.25 \times 10^{-3}\right) \times 170\mathrm{gm}$$

$$=1.0625 \mathrm{gm} .$$
Let, mass of $$AgNO_3$$ sample required = $$m$$ gm.
$$\begin{array}{l}\text { Thus, } m \times \dfrac{85}{100}=1.0625 \\\therefore \quad m=1.0625\times\dfrac{100}{85}\mathrm{gm}=1.25\mathrm{gm}\end{array}$$
So, correct option is (B)
Question 6
1 / -0

A gas cylinder was found unattended in a public place. The investigating team took the collected samples from it. The density of the gas was found to be $$2.380\ g L^{-1}$$ at $$15^oC$$ and pressure. Hence the molar mass of the gas is:

A
30

B
71

C
32

D
58

Solution

Applying ideal gas law,
$$PV = nRT$$
$$P = \dfrac{wt}{Mwt\times Volume}\times RT$$

$$Mwt = \dfrac{d}{P}\times RT = \dfrac{2.380}{1}\times 0.0821\times 288 = 58$$
Question 7
1 / -0

What is the mass of the precipitate formed when $$50$$mL of $$16.9$$% solution of $$AgNO_3$$ is mixed with
$$50$$ mL of $$5.8$$% $$NaCl$$ solution ?
(Ag = $$107.8$$, N =$$14$$, O = $$16$$, Na = $$23$$, $$Cl = 35.5$$)

A
$$7$$ g

B
$$14$$ g

C
$$28$$ g

D
$$3.5$$ g

Solution

$$50mL\,\, 16.9\%\,AgNO_3=\dfrac{50\times 16.9}{100\times 169}=0.5mol\,AgNO_3$$

$$50mL\,\,15.8\%\,NaCl=\dfrac{50\times 5.8}{100\times 5.8}=0.05mol\, NaCl.$$

$$AgNO_3\underset{aq}{+}NaCl\underset{aq}{\rightarrow} AgCl\downarrow +NaNO_3(aq)$$
$$ 0.05$$ $$0.05$$ - -
- - $$0.05$$ $$0.05$$

molecular mass of $$AgCl$$=143.3

mass of ppt$$(AgCl)$$ =$$0.05\times 143.3$$
$$=7.1gm$$

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

3.49g of ammonia at STP occupies a volume of 4.48 $${ dm }^{ 3 }$$ calculate the molar mass of ammonia.

A
13.25 g/mol

B
14.59 $$g/mol$$

C
17.45 $$g/mol$$

D
None of these

Solution

Volume = 4.48 $$dm^3$$ = 4.48 L
Gram Molecular Mass = 22.4 L at STP
So,
4.48 L of ammonia gas weighs = 3.49 g
22.4 L of ammonia weighs = $$\frac{22.4 \times 3.49}{4.48}$$
= 17. 45 g / mol
Question 9
1 / -0

The number of atoms of the $$He$$ in $$104\ amu$$ is:

A
$$3.1 \times 10^{25}$$

B
$$6.2 \times 10^{25}$$

C
$$26$$

D
$$206$$

Solution

Since the mass of one atom of He = 4 a.m.u
Therefore the number of atoms of the He in 104 amu is, $$\dfrac { 104 }{ 4 } =26$$.
Question 10
1 / -0

Pehal and Ishaan were making $$100g$$ sugar solution with concentration by mass $$20\%$$ and $$50\%$$ respectively. Pehal added $$20 g$$ of sugar in her solution whereas ishaan evaporated $$20 g$$ of water from his solution. Now, the solutions were mixed to form a final solution, The concentration by mass of final solution is:

A
$$28\%$$

B
$$45\%$$

C
$$40\%$$

D
$$60\%$$

Solution

For Pehal:- a mass of sugar= $$20g$$
mass of water=$$80g$$
The new mass of sugar=$$20+20=40g$$
$$\therefore$$Concentration of sugar= $$\cfrac { 40 }{ 120 } \times 100=33.33$$%
For Ishaan:- mass of sugar= mass of water= $$50g$$
New mass of water= $$50-20=30g$$
$$\therefore$$ Concentration of sugar= $$\cfrac { 50 }{ 80 } \times 100=62.5$$%
Concentration by mass of final solution= $$\cfrac { New\ mass\ of\ sugar\ in\ Pehal+Mass\ of\ sugar\ in\ Ishaan }{ New\ mass\ of\ solution\ in\ Pehal+New\ mass\ of\ solution\ in\ Ishaan } $$
=$$\cfrac { 40+50 }{ 120+80}$$
=$$45$$%

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Some Basic Concepts of Chemistry Test - 52

Result

Some Basic Concepts of Chemistry Test - 52

Score

-

Rank

-

SHARING IS CARING

Question 1

$$1.4\times 10^{-21}$$ g

$$1.09\times 10^{-21}$$ g

$$5.025\times 10^{23}$$ g

$$16.023\times 10^{23}$$ g

Solution

Question 2

0.9945 g

0.6645 g

0.9375 g

0.8555 g

Solution

Question 3

the same number of atoms

the same number of molecules

more number of moles in flask $$A$$ as compared to flask $$B$$

the same mass of gases

Solution

Question 4

45.23%

15.66%

25.36%

None of these

Solution

Question 5

2.36 g

1.25 g

1.49 g

none of these

Solution

Question 6

30

71

32

58

Solution

Question 7

$$7$$ g

$$14$$ g

$$28$$ g

$$3.5$$ g

Solution

Question 8

13.25 g/mol

14.59 $$g/mol$$

17.45 $$g/mol$$

None of these

Solution

Question 9

$$3.1 \times 10^{25}$$

$$6.2 \times 10^{25}$$

$$26$$

$$206$$

Solution

Question 10

$$28\%$$

$$45\%$$

$$40\%$$

$$60\%$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.