MCQ Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 2

Uncertainty Principle was given by:

0%
Heisenberg
0%
Thomson
0%
Rutherford
0%
Bohr

Explanation

Heisenberg was the scientist who had given the principle of uncertainty. According to his principle position and velocity of an object can not be measured at the same time.

Hence, the correct option is

$A$

Who proposed atom as an indestructible entity?

0%
Rotherford
0%
Dalton
0%
Thomson
0%
Pauli

Explanation

Dalton's Atomic Theory

All matter is made of atoms. Atoms are indivisible and indestructible.
All atoms of a given element are identical in mass and properties
Compounds are formed by a combination of two or more different kinds of atoms.

Hence, option

$B$ is correct.

Determine the mass by mass percentage concentration of a 100g salt solution which contains 20g salt.

0%
30%
0%
20%
0%
80%
0%
none of these

Explanation

Solution:- (B)

$20 \%$

As we know that,

$\text{mass by mass percent} = \cfrac{\text{ Wt. of salt}}{\text{Total wt. of salt}} \times 100$

Given that:-

Weight of salt = 20 g

Total weight of salt = 100 g

$\therefore \text{ mass by mass percent} = \cfrac{20}{100} \times 100 = 20 \%$

Who performed the gold foil experiment?

0%
Thomson
0%
Goldstein
0%
Chadwick
0%
Rutherford

Who is given the credit for creation of first nuclear reactor.

0%
Fermi
0%
Niels Bohr
0%
Einstein
0%
Oppenheimer

Which one of the following scientist's proposed the law of conservation of mass?

0%
J.J. Thomson
0%
John Dalton
0%
Rutherford
0%
Berzelius

Explanation

The law of conservation of mass is given by John Dalton which states that mass can neither be created nor be destroyed in a chemical reaction.
option

$B$ is correct.

According to Indian and Greek philosophers at 400 B.C. ,the continuous sub-division of matter would ultimately yield _______.

0%
atom
0%
electron
0%
proton
0%
neutron

The weight ratio of roasted ore, coke and limestone fed into the blast furnace in the manufacture of cast iron is:

0%
$8:1:4$
0%
$6:4:1$
0%
$8:4:1$
0%
$8:4:3$

$4.6\times 10^{22}$ atoms of an element weight

$13.8$ g. What is the atomic mass of the element?

0%
290 u
0%
180.6 u
0%
34.4 u
0%
104 u

Explanation

$1$ mole of any substance contains

$6.02 \times 10^{23}$ atoms.

Thus,

$4.6 \times 10^{22}$ atoms corresponds to

$\dfrac{4.6 \times 10^{22}}{6.022 \times 10 ^{23}}=0.0764$ moles.

$0.0764$ moles weighs

$13.8$ g.

Thus,

$1$ mole will weigh

$\dfrac{13.8}{0.0764}= 180.6 \ g$ .

Hence, the atomic mass of the element will be 180.6 u.

The isotopes of chlorine with mass numbers

$35$ and

$37$ whose average mass is 35.5 exist in the ratio of:

0%
$1:1$
0%
$3:1$
0%
$3:4$
0%
$3:2$

Explanation

Let the fraction of

$Cl^{35}$ isomer be

$p$ . The fraction of

$Cl^{37}$ isomer will be

$1-p$ .

Average atomic mass

$=\displaystyle {p\times m_{1}+(1-p)m_{2}}$

$\therefore 35.5 = p \times 35 + (1 - p) \times 37$

So,

$p = 0.75$

Hence, the ratio of two isotopes will be

$p: (1-p) = 0.75 : (1-0.75) = 0.75:0.25 = 3:1$

Option B is correct.

The main constituent of LPG is butane. Then
(A) Butane can be liquefied easily under high pressure
(B) Butane is liquefied by chemically reacting with ethane and propane.

0%
Only A is true
0%
Only B is true
0%
Both A and B are true
0%
Both A and B are false

Explanation

$\text{Butane can be liquefied easily under high pressure. Ex: In LPG cylinder.}$

$\text{Burane cannot be liquefied by any chemical reaction.}$

Number of ions present in

$1$ ml of

$0.1$ M barium nitrate solution is:

0%
$1.8\times$ $10^{20}$
0%
$6\times$ $10^{19}$
0%
$1.8\times$ $10^{23}$
0%
$1.8\times$ $10^{-20}$

Explanation

$1$ ml of

$0.1$ M barium nitrate solution corresponds to

$0.001\times 0.1 = 0.0001$ moles.

$Ba{(NO_3)}_2 \longrightarrow Ba^{2+} + {2NO_3}^-$

$1$ molecule of barium nitrate

$Ba(NO_3)_2$ will give

$3$ ions.

Hence, the number of ions present in

$1$ ml of

$0.1$ M

$Ba(NO_3)_2$ solution

$= 0.0001 \times 3 \times 6.02 \times 10^{23}$

$=1.8 \times 10^{20}$ .

A solution is prepared by adding

$2$ g of substance

$A$ to

$18$ g of water. The mass percent of the solute is:

0%
$10$
0%
$20$
0%
$40$
0%
$25$

Explanation

The formula for the mass percentage of solute in solution is

$=\displaystyle \frac{Mass \ of \ solute}{mass \ of \ solution}\times 100$
Substituting values in the above expression, we get
Mass percentage

$=\displaystyle \frac{2}{2+18}\times 100=10$ %

0%
Assertion and reason both are correct statements and reason is correct explanation for assertion.
0%
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
0%
Assertion is correct statement but reason is wrong statement.
0%
Assertion is wrong statement but reason is correct statement.

Which of the following gives the maximum number of ions?

0%
$1$ mole of $K_{4} [Fe(CN)_{6}]$
0%
$1$ mole of $[Co(NH_{3})_{6} ] Cl_{3}$
0%
$1$ mole of $NaCl$
0%
$1$ mole of $K_{3}[Fe(CN)_{6}]$

Explanation

The number of moles in each compound is equal to 1.

Hence, the number of ions in each compound is given by,

$K_4[Fe(CN)_6] \longrightarrow 4K^+ + [{Fe(CN)_6}]^{4-}$

Number of ions = 5

$[Co(NH_3)_6]Cl_3 \longrightarrow [Co(NH_3)_6]^{3+} + 3Cl^-$

Number of ions = 4

$NaCl \longrightarrow Na^+ + Cl^-$

Number of ions = 2

$K_3[Fe(CN)_6] \longrightarrow 3K^+ + [Fe(CN)_6]^{3-}$

Number of ions = 4

Hence, option A is correct.

The number of

$Cl^{-}$ and

$Ca^{2+}$ ions in

$222$ g of

$CaCl_{2}$ is:

0%
$4N_A, 2N_A$
0%
$2N_A, 4N_A$
0%
$1N_A, 2N_A$
0%
$2N_A, 1N_A$

Explanation

$222$ g of

$CaCl_2$ is equal to

$2$ moles of

$CaCl_2$ .

Therefore, it has

$2$ moles of

$Ca^{2+}$ and

$4$ moles of

$Cl^{-}$ .

So, no. of atoms of

$Ca^{2+}$ and

$Cl^{-}$ are

$2N_A$ and

$4N_A$ respectively.

Which of the following is the best example to demonstrate the law of conservation of mass?

0%
12 gm of carbon combines with 32 gm of oxygen to form 44 gm of $CO_{2}$ .
0%
When 72 gm of carbon is heated in a vacuum and no change in its mass takes place.
0%
The weight of a piece of platinum is the same before and after heating in air.
0%
None of these

Explanation

According to the law of conservation of mass:

The total mass of the reactant = the total mass of the product.

Mass of reactants

$= 12 + 32 = 44$ i.e w(C)+ w(oxygen)

Mass of products

$= 44$ i.e. w(carbon dioxide)

Both B and C do not represent a chemical reaction, but only heating of elements, hence they are not the best examples to demonstrate the law of conservation of mass.

In which of the following electrolytes, largest number of ions are present?

0%
$50\:g\:K_{2} SO_{4}\ (mol. wt. = 174)$
0%
$50\:g\:NaCl\ (mol. wt. = 58.5)$
0%
$50\:g\:NH_{4}Cl \ (mol. wt. = 53.5)$
0%
$50\:g\:CaCl_{2\ }(mol. wt. = 111)$

Explanation

The expression for the number of ions is,
Number of ions in the electrolyte

$= \dfrac { \text{Given weight} }{ \text{Molecular weight} } \times N_A \times \text{ number of ions in one molecule}$

Here,

$N_A$ is the Avogadro number.

$K_2SO_{ 4 } \longrightarrow 2K^+ + {SO_4}^{2-}$

It contains

$\dfrac { 50\times 3\times N_A }{ 174 } = 0.86N_A\ ions$

$NaCl \longrightarrow Na^+ + Cl^-$

It contains

$\dfrac { 50\times 2\times N_A }{ 58.5 } = 1.71N_A\ ions$

$NH_{ 4 }Cl \longrightarrow {NH_4}^+ + Cl^-$

It contains

$\dfrac { 50\times 2\times N_A }{ 53.5 } =1.87N_A\ ions$

$CaCl_{ 2 } \longrightarrow Ca^{2+} + 2Cl^-$

It contains

$\dfrac { 50\times 3\times N_A }{ 111 } =1.35N_A\ ions$

Hence, option C is correct.

When 4 L of nitrogen completely reacts with hydrogen, what would be the volume of ammonia gas formed?

0%
$2 L$
0%
$4 L$
0%
$5 L$
0%
$8 L$

Explanation

Reaction:

$N_2+3H_2 \rightarrow 2NH_3$
1 lit 3 lit 2 lit

So, 4 litres of nitrogen will produce 8 litres of

$NH_3$

Option D is correct.

$10.0\ g$ of

$CaCO_{3}$ on heating gave

$4.4\ g$ of

$CO_{2}$ and

$5.6\ g$ of

$CaO$ . This observation is in agreement with the:

0%
law of constant composition
0%
law of multiple proportion
0%
law of reciprocal proportion
0%
law of conservation of mass

Explanation

$CaCO_{3}\rightarrow CaO + CO_{2}$
10 gm 5.6 gm 4.4 gm

Mass of

$CaCO_{3}\: (10\ gm) =$ Mass of

$CaO\ (5.6\ gm)$ + Mass of

$CO_{2}\ (4.4\ gm)$

Hence, it proves the law of conservation of mass.

What will be the ratio of

$Cl^{35}$ and

$Cl^{37}$ respectively in ordinary chlorine if the atomic weight of chlorine is 35.5 ?

0%
1:3
0%
3:1
0%
1:2
0%
2:1

Explanation

Using average molcular mass formula:

$x(35)\quad +\quad (1-x)(37)\quad =\quad 35.5$

$x\quad =\quad 0.75$

$1-x\quad =\quad 0.25$

$\displaystyle \frac { 0.75 }{ 0.25 } =\quad 3:1$

18 carat gold contains ?

0%
18 % gold
0%
24 % gold
0%
75 % gold
0%
60 % gold

Explanation

Caratage is the measurement of purity of gold alloyed with other metals. 24 carat is pure gold with no other metals. Low caratages contain less gold. 18 carat gold contain

$75\%$ of gold &

$25\%$ of other metals often, Copper/Silk.

Active mass of 6% solution of compound X is 2. Molecular weight of X would be :

0%
$6$
0%
$30$
0%
$60$
0%
$90$

Explanation

Active mass of 2 means 2 moles of X in one litre solution.

$\because 100\quad ml\quad solution\quad contains\quad 6\quad gm\quad of\quad X\\ \therefore 1000\quad ml\quad solution\quad contains\quad \frac { 6\times\ 1000 }{ 100 } =\quad 60\quad gm\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 2\quad moles\quad =\quad 60\quad gm\quad X\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 1\quad mole\quad =\quad 30\quad gm\quad X\\ Hence\quad molecular\quad weight\quad of\quad X\quad is\quad 30.$

What is the mass percent of carbon in carbon dioxide?

0%
$0.034$ %
0%
$27.27$ %
0%
$3.4$ %
0%
$28.7$ %

Explanation

Molecular formula of carbon dioxide is

$CO_2$ .
Molar mass of carbon dioxide is

$12+2(16)=44$ g/mol.
Mass percentage of carbon in carbon dioxide is

$\dfrac {12} {44} \times 100 =27.27\%$ .

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. The number molecules does container B have :

0%
N
0%
0.5N
0%
2N
0%
4N

Explanation

1. According to Avogadro's law, the equal volume of gases under a similar condition of pressure and temperature contain equal no. of molecules.

Therefore the number of molecules in container B will also be 2N.

2. As we know, one mole of each gas occupies 22.4-litre volume and one mole of gas contains N

$_A$ molecules at STP.

So at the same temperature and pressure, the equal volume of gas contains equal no of molecules.

Therefore the number of molecules in container B will also be 2N.

Therefore, the option is C.

Gram molar volume for a gas is always considered at ___ conditions.

0%
NTP
0%
STP
0%
variable temperature
0%
none of these

Chemical compounds are formed to have constant composition irrespective of their

0%
Method of formation (or) Source
0%
Conditions for formation
0%
Substances taken for formation
0%
Time of formation

If 25% (w/v) of solution has a density of 1.12 g ml

$^{-1}$ , then:

0%
weight of the solvent in 100 ml solution is 87 g.
0%
weight of the solution per 100 ml is 106 g.
0%
weight of the solution per 100 ml is 112 g.
0%
weight of the solution in 100 ml of solution is 100 g.

Explanation

Density

$= 1.12 g mL^{-1}$

$\Rightarrow$ 1 mL of solution weights 1.12 g

100 ml of solution weights

$100 \times 1.12 =112 g$

25% (W/V) of solution means 25 g of solute is dissolved in 100 ml of solution.

Weight of solvent in 100 ml of solution

$=112 - 25 = 87g$

Barium chloride reacts with sodium sulphate to form Barium sulphate and sodium chloride. Then, according to the law of conservation of mass:

0%
the total mass of reactants, barium chloride, and sodium sulphate taken is greater than the total mass of products, barium sulphate, sodium chloride formed
0%
the total mass of reactants, barium chloride, and sodium sulphate is less than the total mass of products, barium sulphate, and sodium chloride
0%
the total mass of the reactants is equal to the total mass of products
0%
none of the above

A solution is prepared by dissolving 5.64 g of glucose in 60 g of water. Calculate the mass percent of glucose.

0%
8.59%
0%
6.85%
0%
9.34%
0%
3.59%

Explanation

Mass percent of glucose

$=\displaystyle \frac{mass\ of\ glucose}{mass\ of\ solution} \times 100$

$=\displaystyle \frac{5.64}{(5.64 + 60)} \times 100 =$ 8.59%

The volume occupied by half a mole of a gas at STP is:

0%
5.6 L
0%
11.2 L
0%
22.4 L
0%
44.8 L

Explanation

The volume occupied by 1 mole of any gas is 22.4 litres.

So the volume occupied by half a mole of gas would be half of 22.4 litres.

So it will be

$11.2\ litres$ .

Hence, option B is correct.

What is the mass of one atom of

$C-12$ in grams?

0%
1.992 $\displaystyle \times 10^{-23}$ gm
0%
1.989 $\displaystyle \times 10^{-23}$ gm
0%
1.892 $\displaystyle \times 10^{-23}$ gm
0%
1.965 $\displaystyle \times 10^{-23}$ gm

Explanation

Mass of

$1$ mole =

$12\,gm$

Mass of

$6.022 \times 10^{23}\,atom = 12\,gm$

mass of

$1$ atom =

$\dfrac{12}{6.023 \times 10^{23}} = 1.993 \times 10^{-23}\,gm$

A neutral atom of an element has a nucleus with a nuclear charge 13 times and mass 27 times that of hydrogen nucleus. What is the ratio of electrons to protons in its stable positively charged ion?

0%
27 : 13
0%
10 : 19
0%
10 : 13
0%
10 : 14

Explanation

Let the neutral atom of the element be

$X$ .

Given, the nuclear charge of

$X$ is 13 times that of the hydrogen nucleus.

$\Rightarrow$ Atomic number

$=13$

Mass of X is 27 times that of hydrogen nucleus

$\Rightarrow$ Mass number

$=27$ .

Since

$_{13}X^{27}, EC=2, 8, 3$

If X forms stable positively charged ion i.e.,

$X^{+3}$ (as it has 3 valence electrons in its valence shell)

$X^{+3}\rightarrow$ Atomic number

$=13-3=10.$ This implies number of electrons

$=10$ . Number of protons

$=13$ .

$Ratio=10 : 13$

Bromine atom is available in two isotopes,

$^{79}_{35}Br$ (49.7%) and

$^{81}_{35}Br$

$(50.3\%)$ , the average atomic mass of bromine atom is :

0%
$79.016$
0%
$80.076$
0%
$80.006$
0%
$81.016$

Explanation

Solution:- (C)

$80.006$

Percentage of

${^{79}_{35}{Br}} = 49.7 \%$

Percentage of

${^{81}_{35}{Br}} = 50.3 \%$

Therefore,

Average atomic mass

$= \cfrac{49.7 \times 79 + 50.3 \times 81}{100} \\ = \cfrac{3926.3 + 4074.3}{100} \\ = \cfrac{8000.6}{100} = 80.006$

Boron has two stable isotopes,

$^{10}B (19\%)$ and

$^{11}B(81\%)$ . Average atomic weight for boron in the periodic table is:

0%
10.8
0%
10.2
0%
11.2
0%
10.0

Explanation

Average atomic weight

$=\dfrac{ \sum \%abundant \times atomic\ mass}{100}$

$=\dfrac{ 19 \times 10 + 81 \times 11}{100}=10.81$

Option B is the answer.

Which isotope of carbon is used for getting relative atomic masses?

0%
C-12
0%
C-13
0%
C-14
0%
C-15

Explanation

One a.m.u. or one 'u' is equal to

$1.66053892 \times 10^{-24} g$ or

$1.66053892 \times 10^{-27} kg$ . It is equal to

$\dfrac{1}{12}$ of the mass of an atom of carbon-12. It is used as a standard. The masses of all other atoms are determined relative to the mass of an atom of carbon-12.

One a.m.u. or one 'u' is equal to :

0%
$1.66053892 10^{-27} kg$
0%
$1.66053892 10^{-29} kg$
0%
$1.23457656 10^{-27} kg$
0%
$1.23457656 10^{-29} kg$

Explanation

One a.m.u.(atomic mass unit) or one 'u' is

$\dfrac{1}{12}$ of the mass of one carbon-

$12$ atom.
It is equal

$1.6605389210 \times 10^{-24} g$ or

$1.6605389210 \times 10^{-27} kg$ .

The modern atomic mass unit is based on the mass of :

0%
C-12 isotope
0%
hydrogen
0%
oxygen
0%
nitrogen

Explanation

One a.m.u. or one 'u' is equal to

$1.66053892 10^{-24}\ g$ or

$1.66053892 10^{-27}\ kg$ . It is equal to

$\dfrac{1}{12}$ of the mass of an atom of carbon-12.

It is used as a standard. The masses of all other atoms are determined relative to the mass of an atom of carbon-12.

The pair of species having different percentage (by mass) of carbon is :

0%
$CH_{3}COOH$ and $C_{6}H_{12}O_{ 6}$
0%
$CH_{3}COOH$ and $C_{2}H_{5}OH$
0%
$HCOOCH_{3}$ and $HCOOH$
0%
$C_{2}H_{5}OH$ and $CH_{3}OCH_{3}$

Explanation

Hint: Calculate the percentage by mass of carbon in each compound.

Step 1: Calculate the total mass of each molecule:

$CH_3COOH \rightarrow (2\times 12) + (2\times 16) + (4\times 1) = 60\ grams$

$C_6H_{12}O_6 \rightarrow (6\times 12) + (12\times 1) + (6\times 16) = 180\ grams$

$C_2H_5OH \rightarrow (2\times 12) + (6\times 1) + (1\times 16) = 46\ grams$

$HCOOCH_3 \rightarrow (2\times 16) + (4\times 1) + (2\times 12) = 60\ grams$

$HCOOH \rightarrow (2\times 16) + (2\times 12) + (2\times 1)= 46\ grams$

$CH_3OCH_3 \rightarrow (1\times 16) + (2\times 12) + (6\times 1) = 46\ grams$

Step 2: Calculate the mass of carbon in each molecule:

$CH_3COOH \rightarrow 2\times 12=24\ grams$

$C_6H_{12}O_6 \rightarrow 6\times 12 =72\ grams$

$C_2H_5OH \rightarrow 2\times 12 = 24\ grams$

$HCOOCH_3 \rightarrow 2\times 12 = 24\ grams$

$HCOOH \rightarrow 1\times 12= 12\ grams$

$CH_3OCH_3 \rightarrow 2\times 12 = 24\ grams$

Step 3: Calculate the mass percentage of carbon in each molecule:

$CH_3COOH \rightarrow \dfrac{24}{60}\times 100\%= 40\%$

$C_6H_{12}O_6 \rightarrow \dfrac{72}{180}\times 100\%=40\%$

$C_2H_5OH \rightarrow \dfrac{24}{46}\times 100\%=52.17\%$

$HCOOCH_3 \rightarrow \dfrac{24}{60}\times 100\%=40\%$

$HCOOH \rightarrow \dfrac{12}{46}\times 100\%=26.089\%$

$CH_3OCH_3 \rightarrow \dfrac{24}{46}\times 100\%=52.17\%$

Final Answer: Options $B,C$

If the fertilizers listed below are priced according to their nitrogen content, which will be the least expensive per

$50$ kg bag?

0%
Urea, $(NH_2)_2CO$
0%
Ammonia, $NH_3$
0%
Ammonium nitrate, $NH_4NO_3$
0%
Guanidine, $HNC(NH_2)_2$

Explanation

Compound %N
Urea

$(NH_2)_2CO$

$28\times\: \dfrac{100}{60} = 46.67%$

Ammonia

$NH_3$

$14\times \dfrac{100}{17} = 82.35%$

Ammonium nitrate

$NH_4NO_3$

$28\times {100}{80} = 35%$

Guanidine

$HNC(NH_2)_2$

$42\times {100}{59} = 71.18%$
Hence, the least expensive will be Ammonium nitrate

$NH_4NO_3$ .

If an equivalent weight of an element is

$32$ , then the percentage of oxygen in its oxide is :

0%
$16$
0%
$40$
0%
$32$
0%
$20$

Explanation

If an equivalent weight of an element is

$32$ , then it will combine with

$8$ g of oxygen to form metal oxide (from the definition of equivalent weight).
So, % of oxygen

$=8\times \dfrac{100}{40} = 20$

Which of the following has the largest mass of carbon per gram?

0%
$H_2CO_3$
0%
$CH_3CO_2H$
0%
$CH_3OH$
0%
$CH_3CH_2OH$

Explanation

Compound Mass of carbon per gram
(a)

$H_2CO_3 = \dfrac{12}{62}g$ carbon
(b)

$CH_3CO_2H = \dfrac{24}{60}g$ carbon
(c)

$CH_3OH = \dfrac{12}{32}g$ carbon
(d)

$CH_3CH_2OH = \dfrac{24}{40}g$ carbon (maximum)
Hence, largest mass of carbon per gram is for

$CH_3CH_2OH$ .

All the substances listed below are fertilizers that contribute nitrogen to the soil. Which of these is the richest source of nitrogen on a mass percentage basis?

0%
Urea, $(NH_2)_2CO$
0%
Ammonium nitrate, $NH_4NO_3$
0%
Guanidine, $HNC(NH_2)_2$
0%
Ammonia, $NH_3$

Explanation

Compound %N
Urea

$(NH_2)_2CO$

$28\times 100/60 = 46.67\%$
Ammonia

$NH_3$

$14\times 100/17 = 82.35\%$
Ammonium nitrate

$NH_4NO_3$

$28\times 100/80 =35\%$
Guanidine

$HNC(NH_2)_2$

$42\times 100/59 = 71.18\%$
Hence, ammonia is the richest source of nitrogen on a mass percentage basis.

Cost of a fertilizer is based on the percentage of nitrogen in it. Which of the fertilizers should be priced maximum?

0%
$NH_4NO_3$
0%
$NH_2CONH_2$
0%
$KNO_3$
0%
None of the above

Explanation

Molecular weight of

$NH_4NO_3$ is 80,

$NH_2CONH_2$ is 60, and

$KNO_3$ is 101.

Percentage of

$N$ in:

$NH_4NO_3=\displaystyle\dfrac{28\times 100}{80} = 35\%$

$NH_2CONH_2=\displaystyle\frac{28\times 100}{60} = 46.7\%$

$KNO_3=\displaystyle\frac{14}{101}\times 100 = 13.9\%$

Urea, having the highest percentage of nitrogen, should have the highest cost.

Hence, option B is correct.

State true or false.

Out of

$CH_3Cl,$

$C_6H_6Cl_6$ and

$CCl_3CHO$ , maximum percentage of chlorine is in

$C_6H_6Cl_6$ .

0%
True
0%
False

Explanation

Compound % Cl

$C_6H_6Cl_6$

$6\times \dfrac{35.5}{78} + 35.5\times 6 = \dfrac{213}{291} =0.73$

$CCl_3CHO$

$\dfrac{106.5}{147.5} = 0.72$

$CH_3Cl$

$\dfrac{35.5}{50.5} = 0.70$
Hence, maximum chlorine is present in

$C_6H_6Cl_6$ .

$Na_2SO_3\cdot xH_2O$ has

$50\%$

$H_2O$ by mass. Hence,

$x$ is :

0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

Molecular mass of

$Na_2SO_3\cdot xH_2O=126+18x$
Hence,

$\displaystyle\frac{18x}{126+18x}=0.5$ (given)

$\therefore x=7$

Cortisone is a molecular substance containing

$21$ atoms of carbon per molecule. The mass percentage of carbon in cortisone is

$69.98\%$ . It's molar mass is:

0%
$176.5$
0%
$252.2$
0%
$287.6$
0%
$360.1$

Explanation

Let molar mass be

$=M$

Mass of

$21$ carbon atoms

$=252$

$\%$ of carbon

$=\displaystyle\frac{252\times 100}{M}=69.98$

but % of carbon = 69.98 %

$\therefore M=360.1$

Hence, the correct option is

$D$

Which has maximum percentage of

$Cl$ ?

0%
$C_6H_6Cl_6$
0%
$CHCl_3$
0%
$C_6H_5Cl$
0%
$CH_3Cl$

Explanation

Compound % Cl

$C_6H_6Cl_6$

$6\times \dfrac{35.5}{78} + 35.5\times 6 = \dfrac{213}{291} = 73.2\%$

$CHCl_3$

$3\times \dfrac{35.5}{119.5} = \dfrac{106.5}{119.5} = 89.1\%$

$C_6H_5Cl$

$\dfrac{35.5}{112.5} = 31.6\%$

$CH_3Cl$

$\dfrac{35.5}{50.5} = 70.3\%$
Hence, the maximum percentage of chlorine is present in

$CHCl_3$ .

$1$ amu is equal to

$1.66\times 10^{-24}$ g.

0%
True
0%
False

Explanation

The atomic mass unit (amu) is equal to

$\frac{1}{12}$ of the mass of one atom of the carbon-12 isotope.

1 mole carbon-12 atoms weigh 12 g.

$6.022 \times 10^{-23}$ carbon -12 atoms weigh

$12g$ .

1 carbon-12 atom weighs

$\dfrac {12}{6.022 \times 10^{-23}} \ g$

$\therefore 1\ amu = \dfrac{1}{12} \times$ mass of one C-12 atom

$= \dfrac {1}{12}$

$\times \dfrac {12}{6.022 \times 10^{-23}}$

$= 1.66 \times 10^{-24} g$ or

$1.66 \times 10^{-27} kg$ .

A certain metal sulphide,

$MS_2$ , is used extensively as a high temperature lubricant. If

$MS_2$ is

$40.06\%$ by mass of sulphur, metal

$M$ has atomic mass :

0%
$160$ amu
0%
$64$ amu
0%
$40$ amu
0%
$96$ amu

Explanation

Molecular weight of

$MS_2=2S + M$

$\dfrac {64}{M+64} = 0.4006$

$\therefore M=96$

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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