$$a = r \quad b = p \quad c = s \quad d = q$$

$$a = p \quad b = r \quad c = q \quad d = s$$

$$a = r \quad b = q \quad c = s \quad d = p$$

$$a = r \quad b = q \quad c = p \quad d = s$$

MCQ Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 8

Calculate the total number of oxygen atoms present in

$1.80g$ of glucose

$({C}_{6}{H}_{12}{O}_{6})$ .

0%
$46.138\times10^{21}$
0%
$36.138\times10^{21}$
0%
$26.138\times10^{21}$
0%
$6.138\times10^{21}$

Explanation

$Glucose\longrightarrow C_6H_{12}O_6$

$Number\ of\ moles=\dfrac{mass}{molecular\ mass}$

$=\dfrac{1.8}{180}$

$=0.01\ mol$

Now, there are 6 atoms of oxygen

Total number of oxygen atoms

$=$ number of atoms of oxygen

$\times$ number of moles

$\times$ Avagadro's constant

$=6\times0.01\times6.023\times10^{23}$

$=36.138\times10^{21}$

A precipitate of

AgClAgCl and

AgBrAgBr has the mass

0.40660.4066 g. On heating a current of chlorine, the

AgBrAgBr is converted to

AgClAgCl and the mixture loses

0.07250.0725 in mass. The

%% of mass in the original mixture is (as the nearest integer) :

0%
2 %
0%
6 %
0%
5%
0%
95%

Carbon dating method is used to determine the age of ________.

0%
Rock
0%
Fossils
0%
Trees
0%
Ancient monuments

What mass of propene is formed from 34.0 g of iodopropane on heating with ethanolic

$KOH$ , if the yield is 36 %?

0%
17.2 g
0%
8.4 9
0%
3.024 g
0%
1.72 g

Explanation

$CH_3CH_2CH_2I+KOH \longrightarrow CH_3-CH=CH_2+KI+H_2O$

Moles of Iodopropane

$=\cfrac{34}{170}=0.2$

$moles$

$1$

$mole$ of iodopropane gives

$1$

$mole$ propene but yield is

$36\%$ .

So,

$1$

$mole$ of iodopropane will give

$0.36$

$mole$ of propene.

$0.2$

$mole$ of iodopropane will give

$0.2\times 0.36=0.072$

$mole$ of propene.

Mass of propene

$=42\times 0.072=3.024$

$gram$

An organic compound made of

$C, H$ and

$N$ contains

$20$ % nitrogen. Its minimum molecular weight is:

0%
$70$
0%
$140$
0%
$100$
0%
$65$

Explanation

Let the molecular weight be

$X$ . Then

$20$ % of

$X$ is equal to molecular weight of Nitrogen (

$N$ ).

$20$ % of

$X$ =

$14$

$X$ /

$5$ =

$14$

Therefore the minimun molecular weight is

$14×5$ =

$70$ .

$CH_3CH_2CH_2CH_{3(g)}\rightleftharpoons CH_3-\underset{iso-butane}{\underset{CH_3}{\underset{|}{C}H}}-CH_{3(g)}$

0%
$75\%$
0%
$90\%$
0%
$30\%$
0%
$60\%$

Explanation

$K_c$ for the above reaction is represented as

$K_c = \dfrac{isobutane}{butane}$

$(isobutane) = 3(butane)$

So the concentration of sobutane is thrice of butane in 100% solution of the isobutane and butane.

So isobutane

$= 75\%$

What will be the mass percentage of a resulting solution prepared by mixing

$15\%$ (w/w) 500 g aqueous solution of the area with

$25\%$ (w/w) 400g aqueous solution of it?

0%
$18\%$
0%
$20\%$
0%
$25\%$
0%
$15\%$

Explanation

Given

$({w/w})\%$ of

$1^{st}$ solution

$=15$

$\implies 15=\cfrac{500}{Weight\quad of\quad solution(x_1)}\times 100$

$\implies x_1=\cfrac{(500)(100)}{15}$

$\implies x_1=\cfrac{10^{4}}{3}$

Given

$({w/w})\%$ of

$2^{nd}$ solution

$=25$

$\implies 25=\cfrac{400}{Weight\quad of\quad solution(x_2)}\times 100$

$\implies x_2=\cfrac{(400)(100)}{25}$

$\implies x_2=1600$

Total

$({w/w})\%=\left[\cfrac{500+400}{\cfrac{10^4}{3}+1600}\right]\times 100=\cfrac{900}{33.3+16}=\cfrac{900}{49.3}\simeq 18$

$\therefore$ Weight by weight percentage of resulting solution is

$18\%$

The critical volume of a gas is 0.036

$lit. mol^{-1}$ . The radius of the molecule will be (in cm):
(Avogadro Number =

$6 \times 10^{23}$ )

0%
$\displaystyle \left( \frac{9}{4 \pi} \times 10^{-23} \right)^{\dfrac{1}{3}}$
0%
$\displaystyle \left( \frac{8 \pi}{3} \times 10^{-23} \right)^{\dfrac{1}{3}}$
0%
$\displaystyle \left( \frac{3}{8 \pi} \times 10^{-23} \right)^{\dfrac{1}{3}}$
0%
none of these

Explanation

Critical volume of gas=3b [for 1 molecule]

For 1 mole, critical volume=0.036 liters

$\Longrightarrow 0.036\times { 10 }^{ 3 }{ cm }^{ 3 }=3\left( \cfrac { 4 }{ 3 } \pi { r }^{ 3 } \right) \times { N }_{ A } \\ \Longrightarrow r={ \left( \cfrac { 36 }{ 24\pi } \times 10^{ -23 } \right) }^{ \cfrac { 1 }{ 3 } }={ \left( \cfrac { 3 }{ 8\pi } \times 10^{ -23 } \right) }^{ \cfrac { 1 }{ 3 } }$

A sample of oxygen contain

$_{ 8 }^{ 16 }{ O },_{ 8 }^{ 17 }{ O }$ and

$_{ 8 }^{ 18 }{ O }$ atoms has average atomic mass

$16.12$ . If mole

$\%$ of

$_{ 8 }^{ 16 }{ O }$ is

$90$ , then weight

$\%$ of

$_{ 8 }^{ 18 }{ O }$ is

0%
$2$
0%
$2.33$
0%
$1.79$
0%
$8$

$-50^0C$ , liquid

$NH_3$ has ionic product is

$10^{-30}$ . How many amide

$(NH_2{^-})$ ions are present per

$mm^3$ in pure liquid

$NH_3$ ? Take

$N_a \, = \, 6 \times \, 10^{23}$

0%
602
0%
567
0%
450
0%
892

Explanation

$\displaystyle 2NH_3 \to NH_4^++NH_2^-$

The ionic product

$\displaystyle [NH_2^-][NH_4^+]=10^{-30}$

$\displaystyle [NH_2^-]=[NH_4^+]$

$\displaystyle [NH_2^-]^2=10^{-30}$

$\displaystyle [NH_2^-]=10^{-15} \ mol/dm^3$

$\displaystyle 1 \ dm^3 =10^6 \ mm^3$

$\displaystyle [NH_2^-]=10^{-15} \ mol/dm^3 \times \dfrac {1 \ dm^3}{10^6 \ mm^3}$

$\displaystyle [NH_2^-]=10^{-21} \ mol/mm^3$

$\displaystyle N_a=6 \times 10^{23} \ ions/mol$

$\displaystyle [NH_2^-]=10^{-21} \ mol/mm^3 \times \dfrac { 6 \times 10^{23} \ ions} { 1 \ mol }$

$\displaystyle [NH_2^-]= 600 \ ions/mm^3$

Hence,

$\displaystyle [NH_2^-]= 602 \ ions/mm^3$ is correct answer.

What will be the mass of 100 atoms of hydrogen?

0%
$100\ g$
0%
$1.66 \times 10^{-22}$ g
0%
$6.023 \times 10^{23}$ g
0%
$100 \times 6.023 \times 10^{23}$ g

Explanation

Mass of 1 mol or (

$6.022 \times 10^{23}$ atoms) of

$H=1\ g$

Mass of 100 atoms =

$\dfrac{1}{6.022 \times 10^{23}} \times 100$ g

$=0.166 \times 10^{-23}=1.66 \times 10^{-22}$ g

For every ,one

$^{37}Cl$ isotope there are three

$^{35}Cl$ isotopes, in a sample of chlorine. What will be the average atomic mass of chlorine?

0%
35
0%
37
0%
35.5
0%
35.6

Explanation

Average atomic mass of an element existing in different isotopes is given by:

$M_{ avg }=\dfrac { \sum _{ i=1 }^{ n }{ { M }_{ i }{ A }_{ i } } }{ \sum _{ i=1 }^{ n }{ A_{ i } } }$

where

$M_i=$ atomic mass of an isotope with relative abundance of

$A_i$

Given:

$M_1=37 u,A_1=1,M_2=35 u, A_2=3$

on subtitutiing we get:

${ M }_{ avg }=\dfrac { 37\times 1+35\times 3 }{ 1+3 }$

${ M }_{ avg }=35.5\ u$

option C is correct

$\displaystyle\, 80^{\circ}C$ the vapour pressure of pure liquid A is 250 mm of Hg and that of pure liquid B is 1000 mm of Hg. If a solution of A and B boils at

$\displaystyle\, 80^{\circ}C$ and 1 atm pressure, the amount of A in the mixture is (1 atm = 760 mm Hg)

0%
50 mole percent
0%
52 mole percent
0%
34 mole percent
0%
48 mole percent

Explanation

$P$ =

$P°_{a}$ *

$X_{a}$ +

$P°_{b}$ *(

$1$ -

$X_{a}$ )

$760$ =

$250$ *

$X_{a}$ +

$1000$ *(

$1$ -

$X_{a}$ )

$X_{a}$ =

$34$ mole percent

Which symbol is used to represent the unit of atomic mass, amu?

0%
u
0%
A
0%
M
0%
n

4.88 g of

$KClO_3$ when heated produced 1.92 g of

$O_2$ and 2.96 g of KCl. Which of the following statements regarding the experiment is correct?

0%
The result illustrates the law of conservation of mass.
0%
The result illustrates the law of multiple proportions.
0%
The result illustrates the law of constant proportion.
0%
None of the above laws is being illustrated here.

Explanation

$2KClO_3 \rightarrow 2KCl+3O_2$

Mass of the reactant = total mass of the products (

$2.96 + 1.92=4.88\ g$ )

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants. Therefore, this demonstrates the law of conservation of mass.

Option A is correct

Which of the following statements best explains the law of conservation of mass, concerning a chemical change?

0%
100 g of water is heated to give steam.
0%
A sample of $N_2$ gas is heated at constant pressure without any change in mass.
0%
36 g of carbon combines with 32 g of oxygen to form 68 g of $CO_2$ .
0%
10 g of carbon is heated in vacuum without any change in mass.

Explanation

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

Among the given option only option C involves a chemical change/reaction as:

$C+O_2 \rightarrow CO_2$

$m_{reactants}=m_{products}$

$36+32=68\ g$

option C is the best demonstration of the law of conservation of mass.

Oxygen occurs in nature as a mixture of isotopes

$^16O$ ,

$^17O$ and

$^18O$ having atomic masses of 15.995 u, 16.999 u and 17.999 u and relative abundance of 99.763%, 0.037%, and 0.200% respectively. What is the average atomic mass of oxygen?

0%
15.999 u
0%
16.999 u
0%
17.999 u
0%
18.999 u

Explanation

Average atomic mass of an element existing in different isotopes is given by:

$M_{ avg }=\dfrac { \sum _{ i=1 }^{ n }{ { M }_{ i }{ A }_{ i } } }{ \sum _{ i=1 }^{ n }{ A_{ i } } }$

where

$M_i=$ atomic mass of an isotope with relative abundance of

$A_i$

Given:

$M_1=15.995 u,A_1=99.763,M_2=16.999 u, A_2=0.037,M_3= 17.999 u, A_3=0.200$

on subtitutiing we get:

${ M }_{ avg }=\dfrac { 15.995\times 99.763+16.999\times 0.037+17.999\times 0.200 }{ 99.763+0.037+0.200 }$

${ M }_{ avg }=15.999\ u$

option A is correct

The density of a gas is

$1.78 \ g \ L^{-1}$ at STP. The weight of one mole of gas is:

0%
$39.9\ g$
0%
$22.4\ g$
0%
$3.56\ g$
0%
$29\ g$

Explanation

Density

$=1.78 \ g L^{−1}$ at STP

mass of

$1\ L$ gas

$=1.78\ g$

1 mol of gas at STP

$=22.4\ L$

$\therefore\$ mass of

$22.4\ L$ gas

$=22.4 \times 1.78=39.872\ g$

Option

$A$ is correct.

Which of the following gases will have least volume if 10 g of each gas is taken at same temperature and pressure?

0%
$CO_2$
0%
$N_2$
0%
$CH_4$
0%
$HCl$

Explanation

Avogadro’s Law states that under same conditions of temperature and pressure, volume of gases (V) is proportional to the amount of gases in moles(n).

$V \propto n$

$n=\dfrac{mass}{molar\ mass}$

Mass of gases are same = 10 g

$V \propto \dfrac{1}{Molar\ mass}$

Molar mass of

$CO_2=12+2\times 16=44\ g$

Molar mass of

$N_2=2\times 14=28\ g$

Molar mass of

$CH_4=12+4\times 1=16\ g$

Molar mass of

$HCl=1+35.5=36.5\ g$

Order of volume of 10 g of given gases is:

$CH_4>N_2>HCl>CO_2$

Total number of atoms present in 34 g of

$NH_3$ is:

0%
$4 \times 10^{23}$
0%
$4.8 \times 10^{21}$
0%
$2 \times 10^{23}$
0%
$48 \times 10^{23}$

Explanation

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

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

$'X'$ ?

0%
$10\%$
0%
$11.1\%$
0%
$90\%$
0%
$75\%$

Explanation

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=

$\cfrac{5}{5+45} \times 100=10$ %

The correct match is

0%
$a = p \quad b = r \quad c = q \quad d = s$
0%
$a = r \quad b = q \quad c = s \quad d = p$
0%
$a = r \quad b = q \quad c = p \quad d = s$
0%
$a = r \quad b = p \quad c = s \quad d = q$

How much mass of sodium acetate is required to make 250 mL of 0.575 molar aqueous solution?

0%
$11.79 g$
0%
$15.38 g$
0%
$10.81 g$
0%
$25.35 g$

Explanation

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

An impure sample of silver

$(1.5\ g)$ is heated with

$S$ to form

$0.124\ g$ of

$Ag_2S$ . What was the per cent yield of

$Ag_2S$ ?

0%
$21.6\%$
0%
$7.2\%$
0%
$1.7\%$
0%
$24.8\%$

Explanation

Reaction involved:

$2Ag+S \rightarrow Ag_2S$

216 g pure Ag react with 32 g S to give 248 g

$Ag_2S$ .

To obtain 0.124 g

$Ag_2S$ amount of Ag reacts with S is calculated as:

$=\cfrac{216}{248} \times 0.124=0.108$ g

Thus out of 1.5 g sample only 0.108 g (i.e.pure silver) will react to form

$Ag_2S$

% yield of

$Ag_2S=\cfrac{0.108}{1.5} \times 100=7.2$ %

How many number of aluminium ions are present in 0.051 g of aluminium oxide?

0%
$6.023 \times 10^{20}$ ions
0%
$3$ ions
0%
$6.023 \times 10^{23}$ ions
0%
$9$ ions

Explanation

Molar mass of aluminium oxide,

$Al_2O_3=2\times 27 +3 \times 16=102$ g

$Al_2O_3 \quad \longrightarrow \quad 2Al^{3+} \quad + \quad 3O^{2-}$

Number of aluminium ions in

$Al_2O_3$ = 2

Number of ions in 0.051 g of aluminium oxide

$=\dfrac{2 \times 6.022 \times 10^{23}}{102} \times 0.051$

$=6.023 \times 10^{20}$ of Al ions

Hence, option

$A$ is correct.

Which scientist discovered that the same amount of space was occupied by equal numbers of molecules of gases, irrespective of whether it was hydrogen or chlorine or fluorine?

0%
Gay-Lussac
0%
Avogadro
0%
Marconi
0%
Wohler

Explanation

Avogadro first discovered it.

It states that

$1$

$mole$ of any element or compound have

$6.022\times 10^{23}$ atoms or molecules present at any constant temperature and pressure.

What will be the weight of CO having the same number of oxygen atoms as present in 22 g of

$CO_2$ ?

0%
28 g
0%
22 g
0%
44 g
0%
72 g

Explanation

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

$A$

Which of the following statement about Avogadro's hypothesis is correct?

0%
At NTP all gases contain same number of molecules
0%
Under similar conditions of temperature and pressure, gases react with each other in simple ratio.
0%
Under similar conditions of temperature and pressure, equal volumes of all gases contain same number of molecules
0%
Gases always react with gases only at the given temperature and pressure

Read the passage given below and answer the question:
Chemical reactions involve the interaction of atoms and molecules. A large number of atoms/molecules (approximately

$6.023 \times 10^{23}$ ) are present in few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry, and radiochemistry. the following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolyzed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200; 1 faraday =96500 coulombs).
If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is:

0%
200
0%
225
0%
400
0%
446

Explanation

Moles of

$Na^+$ ion

$=2$

$moles$
So, maximum

$2$

$moles$ of

$Na-Hg$ amalgam will be formed.

Weight

$=2\times (23+200)=446$

$gm$

Major problems faced by energy section are

0%
Rising oil prices
0%
Rising oil import bill
0%
Sick SEBs
0%
All of the above

Explanation

$\text { Major problems faced by energy sector are- }$

$\begin{array}{l} \text { - Coal production remain key to energy mix. } \\ \text {- Fourth largest consumer of oil and petrolium } \\ \text { in the world. } \end{array}$

$\begin{array}{l} \text { -Relies on imports to meet growing demand of gas } \\ \text { -electricity shortages hurt industrial output. } \\ \text {- Rising of prices, Rising oil import bill, } \\ \text { sick } S E Bs \text { . } \end{array}$

How many grams

$NaOH$ are present in

$250\ mL$ of

$0.5\ M$

$NaOH$ solution?

0%
$7.32g$
0%
$3.8g$
0%
$5g$
0%
$0.5g$

Explanation

No of moles (n) =

$\dfrac{Molarity \times V_{mL}}{1000}$

Given:

$M = 0.5M$

$V_{mL} = 250 mL$

Therefore,

$n = \dfrac{0.5 \times 250}{1000} = \dfrac{125}{1000}=0.125 mol$

Molar mass of

$NaOH$ =

$40g$

Mass of

$NaOH$ =

$0.125 \times 40 = 5g$

According to Avogadro's hypothesis, equal volumes of gases under the same conditions of temperature and pressure will contain:

0%
the same number of molecules
0%
different number of molecules
0%
the same number of molecules only if their molecular masses are equal
0%
the same number of molecules if their densities are equal.

Explanation

According to avogadro's hypothesis, equal volumes of gases under the same conditions of temperature and pressure will contain, the same no. of molecules.

For eg. One volume of hydrogen combines with one volume of chlorine to produce two volumes of HCl gas.

$H_2\\1vol$

$+$

$Cl_2=\\1 vol$

$2HCl\\2 vol$

'Equal volumes of all gases at the same temperature and pressure contain equal number of molecules'. This law is called as

0%
Boyle's law
0%
Charle's law
0%
Avogadro's law
0%
Gay Lussac's law

Thermal electricity is generated from:

0%
oil
0%
coal
0%
gas
0%
all of the above

A sample of calcium carbonate

$(CaCO_{3})$ has the following percentage composition:

$Ca = 40\%; \ C = 12\%; \ O = 48\%$ .
If the law of constant proportions is true, then the weight of calcium in

$4 g$ of a sample of calcium carbonate from another source will be:

0%
$0.016\ g$
0%
$0.16\ g$
0%
$1.6\ g$
0%
$16\ g$

Explanation

$100$

$g$

$CaCO_3$ ,

$40$

$g$

$Ca$ is present.

$4$

$g$

$CaCO_3$ , Let

$x$

$g$

$Ca$ be present

$\cfrac{x}{4}=\cfrac{40}{100}$

$\implies x=1.6$

$g$

0.46 g of an organic compound was analysed. The increase in mass of

$CaCl_2$ U-Tube was 0.54 g and potash bulb was 0.88 g. The percentage composition of the compound is:

0%
C=52.17%, H=13.04%, O=34.79%
0%
C=50%,H=50%
0%
C=32.19%,H= 18.01%,O=49.8%
0%
C=72%,H=28%

Explanation

Weight of organic compound = 0.46g

Weight of

${H}_{2}O$ = 0.54g

Weight of

$C{O}_{2}$ = 0.88g

$\% \text{ of H/}{H}_{2}O = \cfrac{2}{18} \times \cfrac{0.54}{0.46} \times 100 = 13.04 \%$

$\% \text{ ofC/}C{O}_{2} = \cfrac{12}{44} \times \cfrac{0.88}{0.46} \times 100 = 52.17 \%$

$\% \text{ of O} = 100 - (\% \text{ of C} + \% \text{ of H}) = 100 - (52.17 + 13.04) = 34.79\%$

One gm metal

$M^{3+}$ was discharged by the passage of

$1.81\times 10^{23}$ electrons. What is the atomic mass of metal?

0%
$8g/mol$
0%
$9g/mol$
0%
$10g/mol$
0%
None of these

Explanation

Applying faraday's first law-

$\dfrac{Q}{F} = \dfrac{Wt}{Mwt}\times V.f.$

$\dfrac{ne}{F}= \dfrac{1}{Mwt}\times 3$

$Mwt = \dfrac{1\times F\times 3}{ne}$

$Mwt = 10\dfrac{gm}{mol}$

The number of atoms in 67.2 L of

${ NH }_{ 3 }$ (g) at STP is:

0%
9 ${N }_{ A }$
0%
12 ${N }_{ A }$
0%
3 ${N }_{ A }$
0%
4 ${N }_{ A }$

Explanation

By Avogadro law,

22.4 liter of any gas at STP

$=$ 1mole

$=6.022\times { 1 }0^{ 23 }$ molecules of gas

$=$ NA (Avogadro number)

So, in 67.2 L of

${ NH }_{ 3 }=?$

22.4 L of

${ NH }_{ 3 }=N_A$

1 L of

${ NH }_{ 3 }=\cfrac { NA }{ 22.4 }$

67.2 L of

${ NH }_{ 3 }=\cfrac { 67.2 }{ 22.4 } \times NA$

$=3N_A$ molecules

Now, each molecule contain 4 atoma

Number of atoms

$= 4\times 3N_A = 12N_A$ atoms

The density (in g

$ml^{-1}$ ) of a

$3.60$ M sulphuric acid solution having

$29\%$

$H_2SO_4$ [molar mass

$=98$ g

$mol^{-1}$ ] by mass, will be:

0%
$1.64$
0%
$1.88$
0%
$1.22$
0%
$1.45$

Explanation

Let us consider the problem:

Mass

$%$ % of

${H_2}S{O_4} = 29 \%$

Therefore

$100g$ solution contains

$29g\ {H_2}S{O_4}$

Let the density of solution

$\left( {in{\text{ }}g/ml} \right)$ is

$d$

Molarity of solution

$= \dfrac{{{\text{Moles of}}\ {H_2}S{O_4}}}{{{\text{Volume of solution(ml)}}}} \times 1000$

$\left( {M = 3.60} \right)$

$\dfrac{{\dfrac{{29}}{{98}}}}{{\dfrac{{100}}{d}}} \times 1000 = 3.60$

Hence the answer is

$d = 1.22g/ml$ .

Hence, the correct option is

$C$

A sample of municipal water contains one part of urea (molecular wt

$=60$ ) per million parts of water by weight. The number of urea molecules in a drop of water of volume

$0.05\ ml$ is

0%
$2.5\times 10^{14}$
0%
$5\times 10^{14}$
0%
$5\times 10^{13}$
0%
$5\times 10^{15}$

Explanation

$1 \ ppm = 1 \ mg/L = 10^{-3} g/L$

Water sample contains $1 \ ppm$ urea concentration.

$\therefore \ 10^{-3} \ g$ of urea in $1 \ L$

$x \ g$ urea in $0.05 \times 10^{-3} \ L$

$x= 0.05 \times 10^{-6} \ g$

$60 \ g$ urea $=6.023 \times 10^{23} \ molecules$

$0.05 \times 10^{-6} \ g = n \ molecules$

$n = \cfrac {6.023 \times 10^{23} \times 0.05 \times 10^{-6}}{60}$

$=0.005 \times 10^{17}$

$=5 \times 10^{14} \ molecules$

$10.0g$ a sample of a mixture of calcium chloride and sodium chloride is treated with

$Na_2CO_3$ to precipitate the calcium as calcium carbonate. This

$CaCO_3$ is heated to convert all the calcium to

$CaO$ and the final mass of

$CaO$ is

$1.62gms$ . The

$\%$ by mass of

$CaCl_2$ in the original mixture is :

0%
$15.2\%$
0%
$32.1\%$
0%
$21.8\%$
0%
$11.07\%$

Explanation

$CaCO_3 +heat \longrightarrow CaO +CO_2$

$X$

$1.68gm$

[molecular mass of

$CaCO_3=100u$

molecular mass of

$CaO=56u$

cross multiply through the equation and solve for

$x$ grams of

$CaCO_3$

$x=\dfrac{1.62\times 100}{56}$

$x =2.89gm$

$CaCl_2 +Na_2CO_3 \longrightarrow CaCO_3 +2 NaCl$

Again cross multiply through the equation and solve for

$x$ grams of

$CaCl_2$

$x=2.89$

$x=\dfrac{2.89\times 111}{100}$

$x=3.2gm$

$\%CaCl_2$ in original mixture

$3.2grams$

$CaCl_2$ over

$10gms$ mixture is

$32\%$

If you are given Avogadro's number of atoms of a gas

$X$ . If half of the atoms are converted into

$X_{(g)}^+$ by energy

$\Delta H$ . The IE of

$X$ is :

0%
$\dfrac{2\Delta H}{N_A}$
0%
$\dfrac{2N_A}{\Delta H}$
0%
$\dfrac{\Delta H}{2N_A}$
0%
$\dfrac{N_A}{\Delta H}$

Explanation

Given no. of atoms = Avogadro's no. of atoms =

${N}_{A} = 6.023 \times {10}^{23}$

Given that

$\cfrac{{N}_{A}}{2}$ atoms are ionized, i.e.,

Ionization energy of

$\cfrac{{N}_{A}}{2}$ atoms of gas X =

$\Delta{H}$

$\therefore$ Ionization energy of 1 atom of gas X =

$\cfrac{\Delta{H}}{\left( \cfrac{{N}_{A}}{2} \right)} = \cfrac{2. \Delta{H}}{{N}_{A}}$

Hence, Ionisation energy of gas X is

$\cfrac{2. \Delta{H}}{{N}_{A}}$ .

The mass of a molecule of the compound

$C_{60}H_{122}$ is __________.

0%
$1.4\times 10^{-21}$ g
0%
$1.09\times 10^{-21}$ g
0%
$5.025\times 10^{23}$ g
0%
$16.023\times 10^{23}$ g

Explanation

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.

The total number of electrons in 4.2 g of

$N^{3-}$ ion is: (

$N_A$ is the Avogadro's number )

0%
$2.1 N_A$
0%
$4.2 N_A$
0%
$3 N_A$
0%
$3.2 N_A$

Explanation

${ N }^{ 3- }$ =

$4.2g$ =

$\cfrac { 4.2 }{ 14 } moles$ =

$0.3$ moles

$1$ mole of

${ N }^{ 3- }$ contains

$10N_A$ electrons.

$\therefore$

$0.3$ moles of

${ N }^{ 3- }$ contains

$0.3\times 10 N_A$ electrons.

$3N_A$ electrons

The percentage of pyridine

$(C_5H_5N)$ that forms pyridinium ion

$(C_5H_5NH)$ in a

$0.10$ M aqueous pyridine solution is:

[

$K_b$ for

$C_5H_5N=1.7\times 10^{-9}$ ]

0%
$0.77\%$
0%
$1.6\%$
0%
$0.0060\%$
0%
$0.013\%$

Explanation

We know that

$K_a = \cfrac {K_w}{K_b}= \cfrac {10^{-14}}{1.7 \times 10^{-9}}= 5.88 \times 10^{-6}$

$C_5H_5N+ H_2O \rightleftharpoons C_5H_5NH + C_5H_5NH^+ + OH^-$

Initial conc: $0.1$ $0$ $0$ $0$

Final Conc: $0.1-x$ $x$ $x$

$K_a = \cfrac {x^2}{0.1-x}$ [Since $x <<0.1$ it can be neglected]

$5.88 \times 0.1 \times 10^{-6} =x^2$

$x=7.6 \times 10^{-4}$

% $Pyridine =\cfrac {7.6 \times 10^{-4}}{0.1} \times 100 = 0.77\%$

$\because \left (\cfrac {dissociated \ ions}{concentration \ of \ solution }\times 100\right)$

Hence, the correct option is $A$

The amount of zinc required to produce

$224$ ml of

$H_2$ at NTP on treatment with dilute

$H_2SO_4$ solution will be:

0%
$0.65$ g
0%
$0.065$ g
0%
$65$ g
0%
$6.5$ g

Explanation

Let us consider the problem.

zinc reacts with sulfuric acid according to the following formula.

$Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}$

Therefore,

$1$ mole of zinc reacts to give

$1$ mole of hydrogen

Volume of

$1$ mole of hydrogen at

$STP$ is

$22400$

$ml$

$1$ mole of

$Zn$ is

$65$ grams

Therefore

$0.65$ grams of zinc reacts to give

$224$

$ml$ of hydrogen.

Hence, the correct option is

$A$

What quantity of lime stone on heating will give

$56\ kg$ of

$CaO$ ?

0%
$1000\ kg$
0%
$56\ kg$
0%
$44\ kg$
0%
$100\ kg$

Explanation

$\displaystyle CaCO_3 \xrightarrow {\Delta } CaO+CO_2 \uparrow$

Molar masses of

$\displaystyle CaO$ and

$\displaystyle CaCO_3$ are 56 g/mol and 100 g/mol respectively.

Mass of

$\displaystyle CaO = 56 \ kg = 56000 \ g$

$\displaystyle (\because 1 \ kg = 1000 \ g)$

Number of moles of

$\displaystyle CaO= \dfrac { 56000 \ g}{ 56 \ g/mol}=1000 \ mol$

As per balanced chemical equation, 1 mole of

$\displaystyle CaO$ is obtained by heating 1 mole of

$\displaystyle CaCO_3$ .

Hence, 1000 moles of

$\displaystyle CaO$ will be obtained by heating 1000 moles of

$\displaystyle CaCO_3$ .

Mass of

$\displaystyle CaCO_3 = 1000 \ mol \times 100 \ g/mol=10^5 \ g$

Convert unit from g to kg.

Mass of

$\displaystyle CaCO_3 = 10^5 \ g\times \dfrac {1 \ kg}{1000 \ g}=100 \ kg$

Hence, the correct option is

$D$

What mass of sodium chloride will react with 34.0 g of silver nitrate to produce 17 g of sodium nitrate and 28.70 g of silver chloride if the law of conservation of mass holds good ?

0%
12.35 grams
0%
11.70 grams
0%
9.32 grams
0%
None of these

Explanation

$Nacl + AgNO_{3} \rightarrow NaNO_{3}+Agcl$

By law of conservation of max

total mass on left

By law of conservation of mass

total mass on right

$\Rightarrow x+ 3y = 17+28.7$

$\boxed{x = 11.70 gm}$

1170637_1035310_ans_bc6dc32cebff421cac048a77219c90ef.jpeg

If average molecular wt. of air is 29, then assuming

$N_2$ and

$O_2$ gases are there which options are correct regarding composition of air:
i) 75% by mass of

$N_2$
ii) 75% by moles

$N_2$
iii) 72.41% by mass of

$N_2$

0%
only (i) is correct
0%
only (ii) is correct
0%
both (ii) and (iii) are correct
0%
both (i) and (ii) are correct

Explanation

Average wt. air is

$29$

$Nitrogen=\cfrac{x}{10}\times 29\\x=75\%$

$4.6$

$cm^3$ of methyl alcohol is dissolved in

$25.2$ g of water with density

$0.8$ . Calculate

$\%$ by mass of methyl alcohol.

0%
$14.72$
0%
$10.67$
0%
$12.74$
0%
$13.45$

Explanation

According to he question:

$\% \left( {\dfrac{w}{W}} \right) = \dfrac{{Weight\,of\,solute}}{{Weight\,of\,solute + weight\,of\,solvent\,\left( {water} \right)\,}} \times 100$

Substitutes the values,

$Density = \dfrac{{mass}}{{volume}}$

$mass = 0.8 \times 4.6 = 3.68gm.$

$\% \,by\,mass = \dfrac{{3.68}}{{3.68 + 25.2}} \times 100$

$\% = 12.74\% \,by\,mass$

Hence, option C is correct.

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 8 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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