MCQ Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 7

Referring to the molecular sketch and table above, choose the option that best describes the percentage of carbon in the molecule, in terms of both atomic composition and mass-percent composition.

0%
Carbon makes up approximately $25$ % of the atoms, and $25$ % of the mass.
0%
Carbon makes up approximately $40$ % of the atoms, and $40$ % of the mass.
0%
Carbon makes up approximately $25$ % of the atoms, and $40$ % of the mass.
0%
Carbon makes up approximately $40$ % of the atoms, and $25$ % of the mass.

Explanation

The molecular formula of compound

$= C_4H_9NO_3$

$\%$ of carbon atom

$= \dfrac{4}{4+9+1+3} = \dfrac{4}{17} \times 100 \approx 25\%$

$\%$ of carbon by mass

$= \dfrac{4\times 12}{48+9+14+48} = \dfrac{48}{119} \times 100 = 40.33\%$

A student carries out the following reaction in the laboratory:

Sodium carbonate + Copper sulphate

$\rightarrow$ Copper carbonate + Sodium sulphate

How could you determine the reaction obeyed the law of conservation of mass?

0%
Mass of the sample after the reaction is compared to the mass before the reaction begins
0%
Mass of the sample as the reaction is run is measured
0%
Filtering the reactants
0%
Running the reaction in an open system

Explanation

Law of conservation of mass states that mass of the reactants before the reaction should be same as the mass of the products after the reaction.

We will measure the mass of the product and see if it is the same as the mass of the reactants. Both will be same, thus showing the law of conservation of mass is obeyed.

$\Rightarrow$ Mass of the reactants = Mass of the products after the reaction

Consider the four decompositions shown here:
Reaction I:

$Ca(NO_{3})_{2}(s)\xrightarrow {\Delta} CaO(s) + N_{2}O_{5}(g)$
Reaction II:

$CaSO_{3}(s) \xrightarrow {\Delta} CaO(s) + SO_{2}(g)$
Reaction III:

$Ca(ClO_{3})_{2}(s) \xrightarrow {\Delta} CaCl_{2}(s) + 3O_{2}(g)$
Reaction IV:

$CaCO_{3}(s) \xrightarrow {\Delta} CaO(s) + CO_{2}(g)$
In each case, a solid compound of calcium undergoes thermal decomposition, producing a gaseous product and a calcium containing residual solid.
Assuming equal masses of the four compounds,

$Ca(NO_{3})_{2}, CaSO_{3}, Ca(ClO_{3})_{2}$ , and

$CaCO_{3}$ , are heated in open containers to constant mass, in which case will the mass of solid in the container change by the smallest percentage?

0%
Reaction I
0%
Reaction II
0%
Reaction III
0%
Reaction IV

Explanation

Reaction I:

$Ca_(NO_3)_2$ (s)−→ΔCaO(s)+

$N_2O_5$ (g)Ca(NO3)2(s)→ΔCaO(s)+N2O5(g)

Reaction II:

$CaSO_3$ (s)−→ΔCaO(s)+

$SO_)2$ (g)CaSO3(s)→ΔCaO(s)+SO2(g)

Reaction III:

$Ca(ClO_3)_2$ (s)−→

$ΔCaCl_2$ (s)+

$3O_2$ (g)Ca(ClO3)2(s)→ΔCaCl2(s)+3O2(g)

Reaction IV:

$CaCO_3$ (s)−→ΔCaO(s)+

$CO_2($ g)

The reaction which has less atomic mass at the reatant side and

produced less residual when all reactant has same mass

MASS OF REACTED

$Ca_(NO_3)_2$ = 164

$CaSO_3$ = 120

$Ca(ClO_3)_2$ = 208

$CaCO_3$ = 100

answer is D

Atom is the word derived from the Greek word atomio, which means:

0%
uncutable
0%
cutable
0%
in between
0%
combined

Consider the figure below, in which the blue circles represent atoms of copper metal, atomic mass

$63.5\ amu$ , while the red circles represent atoms in zinc metal, atomic mass

$65.4\ amu$ .
Here are four statements concerning this system. Identify the correct statement or statements.
I. This is a single-phase mixture, with a copper-to-zinc atom ratio of

$3 : 1$ .
II. This is two-phase mixture, with a copper-to-zinc atom ratio of

$8 : 1$ .
III. The mixture is

$80$ % copper, by mass.
IV. This mixture could be an alloy of copper and zinc.

0%
Statement I only.
0%
Statement II only.
0%
Statement I and III only.
0%
Statement I and IV only.

Explanation

Here, no. of atoms of

$Cu = 30\space atoms$

No. of atoms of

$Zn = 10\space atoms$

Single phase mixture is a homogenous mixture with same proportions of atoms.

Here, the ratio of copper to zinc is

$3:1$

So, it is a two phase mixture.

$\Rightarrow \% \space Cu$ by mass

$= \dfrac{3\times 63.5}{3\times 63.5 + 65.4} = \dfrac{190.5}{255.9} = 0.744\times 100 = 74\% \approx 80\%$ by mass

The mixture can be alloy of zinc.

Who arranged elements his periodic table, based upon atomic weight:

0%
Thomson
0%
Mendeleev
0%
Lavoisier
0%
Sir Humphrey Davy

Explanation

Mendeleev arranged the elements first in a regular table known as periodic table according to their molecular masses. The law that the properties of the elements are periodic functions of their atomic numbers.

Also called Mendeleev's law, (originally) the statement that the chemical and physical properties of the elements recur periodically when the elements are arranged in the order of their atomic weights

Hence, the correct option is

$B$

A 5% by mass of an aqueous solution of

$A$ of molar mass 342 g mol

$^{-1}$ is isotonic with 0.878% by mass of an aqueous solution of another substance B. The molar mass of

$B$ in g mol

$^{-1}$ is:

0%
120
0%
180
0%
60
0%
90

Explanation

Isotonic solution has same concentration.

$C_1 = C_2$

$\dfrac {5}{342} = \dfrac{0.878}{\text{Molar mass of B}}$

So $5 \times M=0.878\times 342$

So $M=60.052$

Hence, option C is correct.

Who proposed that equal volumes of gases at the same temperature and pressure have same number of molecules.

0%
Heisenberg
0%
Avogadro
0%
Dalton
0%
Boyle

Explanation

Avogadro's law is: Avogadro's law states that "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules"

Hence, the correct option is $B$

Which one of the following statements is false?

0%
Photochemical smog causes irritation in eyes
0%
London smog is a mixture of smoke and fog
0%
Photochemical smog results in the formation of PAN
0%
London smog is oxidising in nature

Who supported the Four Element Concept of Empedocles?

0%
Dalton
0%
Avogadro
0%
Einstein
0%
Aristotle

Radioactivity is the caused by the disintegration of:

0%
Atom
0%
Ion
0%
Molecule
0%
Nucleus

A compound

$X_2O_3$ contains 31.58% oxygen by weight. The atomic weight of

$X$ is:

0%
$344.66 g/mol$
0%
$52.00 g/mol$
0%
$45.01 g/mol$
0%
$104.0 g/mol$

Explanation

Let

$x$ g/mol be the atomic weight of

$X$ . The atomic weight of

$\displaystyle O = 16$ g/mol. The molecular weight of

$\displaystyle X_2O_3 = 2(x)+3(16)=2x+48$ g/mol

Weight percent of

$\displaystyle O = \dfrac {3(16)}{2x+48} \times 100= 31.58$

$\displaystyle \dfrac {48}{2x+48} = 0.3158$

$\displaystyle 2x+48 = \dfrac {48}{0.3158}$

$\displaystyle 2x+48=151.9$

$\displaystyle 2x=103.9$

$\displaystyle x=52.00$ g/mol.

Hence, the atomic weight of

$X$ is

$52.00 g/mol.$

So, option (B) is the correct answer.

At room temperature and pressure, two flasks of equal volumes are filled with

$H_2$ and

$SO_2$ separately. Particles which are equal in number, in the two flasks are:

0%
atoms
0%
electrons
0%
molecules
0%
neutrons

A sample of potato was ground in a ball mill to give a starch-like molecule of lower molecular weight. the product analysed 0.086% phosphorous if each molecule is assumed to contain one atom of phosphorous, what is the molecule weight of material

0%
36000
0%
37000
0%
46000
0%
3600

Explanation

the percentage is given question is .86%.therefore

in 100 gm of the starch molecule will contain .86 gm of phosphorus

given 1atom of phosphorus is contained in one molecule.

therefore Na(avagadro no) of phosphorus will also contain Na molecule or 1mole which corresponds to molecular mass.

0.86 gm of phosphorus equal to 100 gm of a molecule

31 gm corresponds to

$\dfrac{100\times31}{0.086}$ =36045gm

molecular weight is 36045 gm

How many tons of

$Zn$ can be obtained form 10 tons of

$ZnS$ ?

0%
5.025
0%
10.487
0%
15.282
0%
None of these

Hydrogen combines with chlorine to form HCI. It also combines with sodium to form NaH. If sodium and chlorine also combine with each other, they will do so in the ratio of their masses as:

0%
23 : 35.5
0%
35.5 :23
0%
1 : 1
0%
23 : 1

Explanation

Since, mass ratio of

$H : Cl = 1 : 35.5$ in HCl and

$Na : H = 23 : 1$ in NaH.
And since according to law of constant proportion compounds always contains its elements in a fixed ratio by mass.
So, for

$\text{ 1gm of H}$ we are having

$\text{ 35.5gm Cl and 23gm Na.}$
Hence, the ratio

$Na : Cl = 23 : 35.5$ .
Hence, the answer is option A.

$SO_2$ gas was prepared by
(i) burning sulphur in oxygen,
(ii) reacting sodium sulphite with dilute

$H_2SO_4$ and
(iii) heating copper with

$conc. H_2SO_4$ . It was found that in each case sulphur and oxygen combined in the ratio of 1 :The data illustrates the law of :

0%
conservation of mass
0%
multiple proportions
0%
constant proportions
0%
reciprocal proportions

Explanation

Ratio of sulfur & oxygen

$=1:1.$

Thus illustrates law of constant proportions because the ratio is a constant.

When

$22.4\ L$ of

$C_{4}H_{8}$ at

$STP$ is burnt completely,

$89.6\ L$ of

$CO_{2}$ gas the

$STP$ and

$72\ g$ of water are produced. The volume of the oxygen gas the STP consumed in the reaction is closest to

0%
$89.6\ L$
0%
$112\ L$
0%
$134.4\ L$
0%
$22.4\ L$

Explanation

$\underset {\underset {At S.T.P.}{22.4\ lit}}{C_{4}H_{8}} + 6O_{2} \rightarrow 4CO_{2} + \underset {\underset {at S.T.P.}{89.6\ lit.\ 72\ g}}{4H_{2}O}$

$1\ mole \ 4\ mole \ \dfrac {72}{18} = 4\ mole$

For complete combustion of

$1\ mole C_{4}H_{8}$

$6\ mole\ O_{2}$ required

$n_{O_{2}} = 6\ mole$

$V_{O_{2}} = 6\times 22.4$

$V_{O_{2}} = 134.4\ lit.$

Hence, the correct option is

$\text{C}$

The number of g-atom of oxygen in

$6.02\times { 10 }^{ 24 }$

$CO$ molecule is

0%
$1$
0%
$0.5$
0%
$5$
0%
$10$

Explanation

One molecule of

$CO$ contains one oxygen atom

$\therefore 6.02\times { 10 }^{ 24 }\quad$ molecules of

$CO$ contain

$6.02\times { 10 }^{ 24 }$ oxygen atoms

$6.02\times { 10 }^{ 23 }$ atoms of oxygen

$\equiv$

$1g$ atom of oxygen

$\therefore 6.02\times { 10 }^{ 24 }$ atoms of oxygen

$\equiv$

$10g$ atom of oxygen.

A mixture of calcium and magnesium carbonates weighing 1.4 g was strongly heated until no further loss of weight was perceived. The residue weighed 0.76 g. What percentage of

$MgCO_3$ was present in the mixture?

0%
20.45
0%
18.55
0%
16.33
0%
22.40

Explanation

$CaCO_3 + MgCO_3 \rightarrow CaO + MgO$

let mass of

$CaCO_3$ = x g

and mass of

$MgCO_3$ =

$1.4-x$ g

moles of

$CaCO_3$ =

$\dfrac{x}{100}$ = moles

$CaO$ produced

moles of

$MgCO_3$ =

$\dfrac{x}{84}$ = moles

$MgO$ produced

mass of

$CaO$ produced = moles * no. of moles =

$\dfrac{(x) * 56}{100}$

mass of

$MgO$ produced = moles * no. of moles =

$\dfrac{(1.4-x) * 40}{100}$

so, total weight of product =

$\dfrac{(1.4-x) * 56}{100}$ +

$\dfrac{(1.4-x) * 40}{100}$ = 0.76

so, value of x = 1.356 g

percentage of

$CaCO_3$ =

$\dfrac{1.356}{1.4} * 100$ = 96.8 %

percentage of

$MgCO_3$ = 3.14 %

Which of the following is/are correct?

0%
One atomic mass unit is a mass unit equal to exactly one -twelfth (1/12th) the mass of one atom of carbon-12.
0%
One atomic mass unit is a mass unit equal to exactly one - sixteenth (1/16th) the mass of one atom of oxygen-16.
0%
The relative atomic mass of the atom of an element is defined as the average mass of the atom, as compared to 1/12h the mass of one carbon-12 atom.
0%
None of the above.

2.0 g mixture of sodium carbonate and sodium bicarbonate on heating to constant mass gave 224 mL of

$CO_2$ at NTP. The % mass of sodium bicarbonate in the mixture is:

0%
50
0%
54
0%
80
0%
84

Explanation

Sodium bicarbonate decomposes in

the following manner.

$2NaHCO_{3}\rightarrow Na_{2}CO_{3}+H_{2}O+CO_{2}$

Sodium Carbonate is stable and

doesn't dissociate upon normal

heating.

Vol. of

$CO_{2} = 224\,ml$

$22.4\,Ltr = 1\,mole$

So moles of

$CO_{2} = 0.224/22.4 = 0.01\,mole$

$0.01\,mole = 44\times 0.01\,g = 0.44\,g$

1 mole

$CO_{2}$ is procured by 2 mole

$NaHCO_{3}$

$\therefore$ Moles of

$NaHCO_{3} = 2\times 0.01 = 0.02$

$= 0.02\times 84 = 1.689$

$\%$ age

$= \dfrac{mass}{Total\,mass}\times 100 = \dfrac{1.68}{2}\times 100 = 84\%$

1133200_665393_ans_8fd7b27a856d4b7f9f7d561ac5882102.jpg

How many grams of

$NaOH$ will be required to prepare

$500\ g$ solution containing

$10$ %

$\dfrac {w}{w}NaOH$ solution?

0%
$100\ g$
0%
$50\ g$
0%
$0.5\ g$
0%
$5.0\ g$

Explanation

$500\ g$ of solution of 10% w/w

$NaOH$ present.

Mass of the solute

$= \dfrac {10\times 500}{100} = 50\ g$

So,

$50\ g$ of

$NaOH$ will be required to prepare

$500\ g$ of

$10$ %

$\dfrac {w}{w}\ NaOH$ solution.

$5.0\ g$ sample of urea when heated with

$NaOBr$ and

$NaOH$ gives

$1120\ ml$ of nitrogen at STP. The percentage purity of the sample is:

0%
$50$ %
0%
$40$ %
0%
$60$ %
0%
$30$ %

Explanation

${ NH }_{ 2 }CO{ NH }_{ 2 }+3NaOBr+2NaOH\longrightarrow { N }_{ 2 }+3NaBr+{ Na }_{ 2 }{ CO }_{ 3 }+3{ H }_{ 2 }O$

$\because$

$22400\ ml$ of

${N}_{2}$ is obtained from

$60\ g$ of urea at S.T.P.

$\therefore$

$1120\ ml$ of

${N}_{2}$ will be obtained from

$\Rightarrow \cfrac { 60\times 1120 }{ 22400 } g$ of urea

$=3\ g$ or urea

Now,

$\because$

$5\ g$ sample contains

$3.0\ g$ of urea

$\therefore$

$100g$ sample will contain

$\cfrac { 3.0\times 100 }{ 5.0 } =60g$ of urea

Hence, the purity of urea sample is

$60$ %

The molecular weights of

$O_2$ and

$N_2$ are

$32$ and

$28$ respectively. At

$15^0$ C, the pressure of

$1$ gm

$O_2$ will be the same as that of

$1$ gm

$N_2$ in the same bottle at the temperature:

0%
$-21^0$ C
0%
$-13^0$ C
0%
$15^0$ C
0%
$56.4^0$ V

Explanation

Temperature of

$O_{2}$

$=273+15=298 K$

Temperature of

$N_{2}$

$=x$

$n$

$\propto$

$\dfrac 1T$ , as pressure is constant.

$\Longrightarrow { n }_{ 1 }{ T }_{ 1 }={ n }_{ 2 }{ T }_{ 2 }\Longrightarrow \left( \cfrac { 1 }{ 32 } \right) \left( 298 \right) =\left( \cfrac { 1 }{ 28 } \right) \left( x \right) \\ \Longrightarrow x=\cfrac { 28 }{ 32 } \times 298=260.75K$

Temperature in

$^oC=260.75-273=-13 ^oC$

Hence, option

$B$ is correct.

The maximum amount of

$Ba{SO}_{4}$ precipitated on mixing

$20mL$ of

$0.5M$

$Ba{Cl}_{2}$ with

$20mL$ of

$1M$

${H}_{2}{SO}_{4}$ is?

0%
$0.25$ mole
0%
$0.5$ mole
0%
$1$ mole
0%
$0.01$ mole

Explanation

$\textbf{Step 1: Calculation for number of moles of}$

$BaCl_2\ \textbf{and} \ H_2SO_4$ .

$\text{Molarity of }BaCl_{2}=0.5M$

$\text{Volume of }BaCl_{2}=20mL$

Number of moles

$= \cfrac{Molarity\times volume\ (in\ mL)}{1000}$

$\text{Number of moles of }BaCl_{2}=\dfrac{0.5\times 20}{1000}=0.01moles$

$\text{Molarity of }H_{2}SO_{4}=1M$

$\text{Volume of }H_{2}SO_{4}=20mL$

$\text{Number of moles of }H_{2}SO_{4}=\dfrac{1\times20}{1000}=0.02moles$

$\textbf{Step 2: Calculation of the number of moles of}$

$BaSO_4\ \textbf{precipitated}$

Reaction of

$BaC{l_2}$ and

${H_2}S{O_4}$ :

$BaC{l_2} + {H_2}S{O_4} \to BaS{O_4}\left( \downarrow \right) + 2HCl$

From stoichiometry,

$1$ mole of

$BaC{l_2}$ reacts with

$1$ mole of

${H_2}S{O_4}$ .

$\Rightarrow 0.01$ moles of

$BaC{l_2}$ will react only with

$0.01$ moles of

${H_2}S{O_4}$ .

So,

$BaC{l_2}$ is the limiting reagent.

Again from the reaction,

$0.01$ moles of

$BaC{l_2}$ will react with

$0.01$ moles of

${H_2}S{O_4}$ to give

$0.01$ moles of

$BaS{O_4}$ and

$0.02$ moles of

$HCl$ .

Hence, the correct answer is option D.

The number of gram molecule of oxygen in

$6.02 \times 10^{24}$ CO molecules is :

0%
10 gm molecules
0%
5 gm molecules
0%
1 gm molecules
0%
0.5 gm molecules

Explanation

Given :

$6.02\times 10^{24}$

$CO$ molecules

We know,

$1$ mole

$CO$ equivalent to

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

$CO$

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

$CO$ contain

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

$O$

Then,

$6.02\times 10^{24}$

$CO$ molecules contain

$6.02\times 10^{24}$ atoms of

$O$

$\Rightarrow$ no. of gram atoms of

$O=\cfrac {6.02\times 10^{24}}{6.02\times 10^{23}}$

$=10gm$ atoms

$\longrightarrow 1$

As oxygen is a diatomic molecule.

No. of gm of molecules of oxygen=

$\cfrac {10gm\quad atoms}{2\quad atoms/molecule}=5gm$ molecules

Therefore, the correct option is B.

What is the relative molecular mass of sodium sulphate? (The relative atomic masses of

$Na, S$ and

$O$ are

$23 u, 32 u$ and

$16 u$ respectively)

0%
140
0%
142
0%
146
0%
148

Explanation

Sodium sulphate is

$Na_2SO_4$ . Its relative molecular mass is

$2(23)+32+4(16) = 142 \: u$

Calculate the silver ion concentration in

$0.2\ M$ solution as

$Ag(NH_{3})_{2}NO_{3} \cdot K_{silver} = 32\times 10^{-9}$ .

0%
$4\times 10^{-2}M$
0%
$2\times 10^{-3}M$
0%
$3\times 10^{-3}M$
0%
$5\times 10^{-2}M$

Explanation

The reaction involved is:

$[Ag(NH_3)_2]NO_3+H^+\longrightarrow Ag^++2NH_3+H^++{NO_3}^-$

$\Rightarrow K_p=\cfrac {[Ag][NH_3]^2[{NO_3}^-]}{[Ag(NH_3)_2]NO_3}$

$\Rightarrow 32\times 10^{-9}=\cfrac {[Ag]^4}{0.2}$

$[\because K_{og}=32\times 10^{-9}$ given]

[complex]=

$0.2M$ given

$\Rightarrow Ag= 2\sqrt {2}\times 10^ {-2.5}$

$=2.8\times 10^{-2.5}\approx 3\times 10^{-3} M$

The percentage composition of

$Cr$ in

${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$ is (Given molar mass of

${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }=294.0g$

$mol$ )

0%
$60.84$
0%
$28.54$
0%
$51.3$
0%
$35.37$

Explanation

Given,

$K_2Cr_2O_7$

Molar mass of

$K_2Cr_2O_7=294$ g/mol

$\longrightarrow (1)$

Mass percentage=

$\cfrac {Mass\quad of\quad required\quad element\quad in\quad total\quad sample}{Mass\quad of\quad total\quad sample}\times 100$

or mass percentage means, the amount of required element present in

$100g$ of sample.

From, (1)

$1$ mol i.e;

$294g$

$K_2Cr_2O_7$ contains

$78g$

$K$ ,

$104g$

$Cr$ &

$112$

$O$

Now, if

$294g$

$K_2Cr_2O_7$ contains

$104g$

$Cr$ then

$100g$ sample of

$K_2Cr_2O_7$ would contain:

$\cfrac {100\times104}{294}=35.374g$ of

$Cr$

$\therefore$ Mass percentage of

$Cr$ in

$K_2Cr_2O_7=35.37$ %

A solution has

$0.05\ M - Mg^{+2}$ and

$0.05M - NH_{3}$ . Calculate the minimum concentration of

$NH_{4}Cl$ required to prevent only precipitation of

$Mg$ as

$Mg(OH)_{2}\cdot K_{sp}$ of

$Mg(OH)_{2} = 9\times 10^{-12}$ and

$K_{b}$ of

$NH_{3} = 1.8\times 10^{-5}$ .

0%
$0.057\ M$
0%
$0.067\ M$
0%
$0.045\ M$
0%
$1\ M$

The concentration of

$OH^-$ ions in a solution with the

$H^+$ ions concentration of

$1.3\times 10^{-4}$ M is?

0%
$7.7\times 10^{-4}$
0%
$1.3\times 10^{-4}$ M
0%
$2.6\times 10^{-8}$ M
0%
$7.7\times 10^{-11}$ M

Explanation

Product of hydrogen ion concentration and hydroxide ion concentration is equal to 1.0 x 10

$^{-14}$ .

${[OH}^-] \times [H^+] = 1\times10^{-14}$ .

${[OH}^-]$ =

$\dfrac{10^{-14}} {1.3×10^{−4}} = 7.7\times10^{-11}$ .

Thus, the correct answer is option D.

In carbon disulphide

$(CS_{4})$ , the mass of sulphur in combination with

$3.0\ g$ of carbon is

0%
$4.0\ g$
0%
$6.0\ g$
0%
$64.0\ g$
0%
$16.0\ g$

Explanation

Given, Carbon disulphide

$CS_2$

Molar mass of

$CS_2=12g+32\times 2g$

$=12+64g=76g$

i.e;

$12g$ of

$C$ reacts with

$64g$ of

$S$ to form

$76g$

$CS_2$

Also, in

$12g$ of

$C$ forms

$76g$

$CS_2$

$\Rightarrow 3.0g$ of

$C$ forms

$\cfrac {76\times3}{12}g$ of

$CS_2$

$\Rightarrow 19g$ of

$CS_2$ is formed.

Also,

$76g$

$CS_2$ has

$64g$ of

$S$

then,

$19g$

$CS_2$ will have

$\cfrac {19\times 64}{76}g$ of

$S$

$\Rightarrow 16.0g$ of

$S$ is present.

If 100 gm of salt solution contains 20 g of salt dissolved in it, the percentage of the mass of the solution is________.

0%
$10\%$
0%
$20\%$
0%
$15\%$
0%
$25\%$

Explanation

Given,

$100gm$ solution contains

$20g$ salt in it

$\therefore$ Percentage of mass in the solution=

$\cfrac {Mass\quad of\quad salt\quad dissolved}{Mass\quad of\quad solution}\times 100$

$=\cfrac {20gm}{100gm}\times 100$

$=20$ %

Decreasing order of atomic weight is correct of the elements given below?

0%
$Fe > Co > Ni$
0%
$Ni > Co > Fe$
0%
$Co > Ni > Fe$
0%
$Co > Fe > Ni$

Explanation

Atomic weight of Iron

$\approx 56$ u

Atomic weight of Cobalt

$\approx 59$ u

Atomic weight of Nickel

$\approx 58.7$ u

So, the order of atomic weights is

$Co>Ni>Fe$ .
Option

$C$ is correct.

300 gm of 25% w/w solution of solute A is mixed with 400 gm of 40 (w/w) solution of another solute B. What is the w/w percentage of new mixture.

0%
10.71% in A
0%
35% in B
0%
25% in A
0%
40% in B

Explanation

A : 300 g of 25% w/w
B : 400 g of 40% w/w

$w/w \, \% \, of \, A= \dfrac{25\% \, of \, 300}{300 + 400} = \dfrac{75}{700} \times 100 = 10.71\%$

$w/w \, \% \, of \, B = \dfrac{40\% \, of \, 400}{300 + 400} \times 100 = 22.85\%$

A solution containing 4.2 g of KOH and

$Ca(OH)_2$ is neutralized by an acid. If it consumes 0.1 equivalent of acid, the % of KOH in the original sample is:

0%
15%
0%
25%
0%
35%
0%
45%

Explanation

Let the amount of KOH in the solution be x, then amount of

$CaOH_2$ is (4.2-x)

Equivalent mass of KOH =

$\dfrac{56}{1} = 56$

Equivalent mass of

$Ca(OH)_2 =\dfrac{74}{2} = 37$

Gram equivalents of KOH + Gram equivalents of

$Ca(OH)_2$ = Gram equivalents of the acid

$\Rightarrow \dfrac{x}{56}+\dfrac{4.2-x}{37}=0.1$

$\Rightarrow 37x + 235.2 - 56x = 207.2$

$\Rightarrow 235.2-207.2=56x-37x$

$\Rightarrow 28 = 19x$

$\Rightarrow x=\dfrac{28}{19} = 1.47$

Therefore, amount of KOH in the solution = 1.47

Percentage of KOH =

$\dfrac{1.47}{42} \times 100 = 35$

Hence, Option C is correct.

Sodium reacts with excess oxygen to form sodium oxide.A student wants to prepare 1.24 g of sodium oxide.While doing the calculations, he uses atomic number of sodium instead of obtained due to this mistake?

0%
$11\%$
0%
$23\%$
0%
$48\%$
0%
$60\%$

The weight of mixture containing HCl and

$H_2SO_4$ is

$0.1$ g on treatment with an excess of an

$AgNO_3$ solution, reacted with this acid mixture gives

$0.1435$ g of AgCl, weight percentage of

$H_2SO_4$ in mixture is:

0%
$36.5$
0%
$63.5$
0%
$50$
0%
$80$

Explanation

Let mass of

$HCl$ be

$x$

$\therefore$ Mass of

$H_2SO_4$ is

$0.1-x$

Moles of

$HCl=\cfrac{x}{36.5}$

$HCl+AgNO_3\longrightarrow AgCl\downarrow +HNO_3$

Moles of

$HCl=$ moles of

$AgCl$

Moles of

$AgCl=\cfrac{0.1435}{143.3}=10^{-3}$ moles

$\therefore \cfrac{x}{36.5}=10^{-3}$

$x=36.5\times 10^{-3}$

Weight of

$HCl=0.0365$

$g$

Weight of

$H_2SO_4=0.0635$

$\%$ weight of

$H_2SO_4=\cfrac{0.0635}{0.1}\times 100=63.5\%$

Two organic compounds

$\underline {R}$ and

$\underline {S}$ , both containing only

$C$ and

$H$ yield on analysis, the same percentage composition by weight.

$C = (12/ 13) \times 100$ % and

$H = (1/13)\times 100$ %.

$\underline {R}$ decolourises

$Br_{2}$ - water but

$\underline {S}$ does not.

0%
$C_{2}H_{2}, C_{6}H_{6}$
0%
$C_{6}H_{6}, C_{2}H_{2}$
0%
$C_{2}H_{4}, C_{2}H_{6}$
0%
$C_{2}H_{2}, C_{3}H_{8}$

Explanation

According to question the compound have equal % composition of mole. So, component carbon

$(C)$ and the component

$H$ have same in number.

$\rightarrow \quad C_n H_n$ type of compound

Now,

According to bromine water reaction see in fig.

But as this is question of physical chemistry, we know the

$C_n H_n$ type of molecules should be obtained but in options only

$(A)$ option has only

$C_n H_n$ like atoms.

$(A)$ is correct answer.

A mixture contains equimolar quantities of carbonates of two bivalent metals. One metal is present to the extent of

$13.5\%$ by weight in the mixture and

$2.58$ g of the mixture on heating leaves a residue of

$1.35$ g. What percentage by weight of the other metal is there?

0%
$31.6$
0%
$21.3$
0%
$19.2$
0%
$17.8$

Explanation

Let equimolar metal carbonates have mole be

$n$

Let Metal be

$M_1$ and

$M_2$

${ M }_{ 1 }{ CO }_{ 3 }\underrightarrow { \Delta } { M }_{ 1 }O+{ CO }_{ { 2 }_{ (g) } }\uparrow$

${ M }_{ 2 }{ CO }_{ 3 }\underrightarrow { \Delta } { M }_{ 2 }O+{ CO }_{ { 2 }_{ (g) } }\uparrow$

Let molecular wt. of

${ M }_{ 1 }\rightarrow { w }_{ 1 }$

molecular wt. of

${ M }_{ 2 }\rightarrow { w }_{ 2 }$

for % composition of metal

${ M }_{ 1 }=\cfrac { n{ w }_{ 1 } }{ n\left( { w }_{ 1 }+60 \right) +n\left( { w }_{ 2 }+60 \right) } =0.135\quad -(1)$

$\Rightarrow$ we have

$n\left( { w }_{ 1 }+60 \right) +n\left( { w }_{ 2 }+60 \right) =2.58gm\quad -(2)$

$CO_2$ librated

$=2.58-1.35=1.23gm$

$\therefore$ wt. of

$CO_2$ in mixture

$=n(44)+n(44)$

$=88n\quad gm$

$88n\quad gm=1.23gm$

$n=\cfrac {1.23}{88} =0.01397$

Put

$n=0.01397$ mol in

$(1)$ we get

$\cfrac {0.01397 w_1}{2.58} =0.135$

$[w_1 =24.93gm]$

From

$(2)$

$n(w_2 +w_1 +120)=2.58$

$0.01397(24.93+ w_2 +120)=2.58$

$2.0246+0.01397 w_2 =2.58$

$[w_2 =39.75gm]$

% composition of

$M_2 =\cfrac { n{ w }_{ 2 } }{ 2.58 } =\cfrac { 0.01397\times 39.75 }{ 2.58 } =0.215$

$=21.5$ %

$\text{0.80 g}$ of impure

$(NH_4)_2SO_4$ was boiled with

$100 ml$ . of a

$\text{0.2 N NaOH}$ solution till all the

$NH_3(g)$ evolved.The remaining solution was diluted to

$\text{250 mL.25 mL}$ of this solution was neutralized using

$5 mL$ of a

$\text{0.2 n H_2SO_4}$ solution.The percentage purity of the

$(NH_4)_2SO_4$ sample is

0%
$82.5$
0%
$72.5$
0%
$62.5$
0%
$17.5$

Explanation

Total m-eq. of

$NaOH$ taken

$=20$
m-eq. of

$H_2SO_4$ = m-eq. of

$NaOH$ reacted

$= \dfrac{5 \times 0.2}{25} \times 250 = 10$

m-eq. of

$NaOH$ reacted

$= 20 - 10 = 10$

$(NH_4)_2SO_4 + 2NaOH$

$\Rightarrow Na_2SO_4 + 2NH_3 + 2H_2O$
m-moles of

$(NH_4)_2SO_4$ reacted

$= 5$
wt. of

$(NH_4)_2SO_4$ in sample

$= \dfrac{0.66}{0.80} \times 100 = 82.5$

One mole of any substance contains

$6.022\times 10^{23}$ atoms/molecules. Number of molecules of

$H_2SO_4$ present in

$100$ mL of

$0.02$ M

$H_2SO_4$ solution is __________.

0%
$12.044\times 10^{20}$ molecules
0%
$6.022\times 10^{23}$ molecules
0%
$1\times 10^{23}$ molecules
0%
$12.044\times 10^{23}$ molecules

Explanation

Moles of

$H_2SO_4=$ Molarity of

$H_2SO_4\times$ Volume of solution

$(L)$

$=0.02\times 0.1$

$=2\times 10^{-3}$ moles

No. of

$H_2SO_4$ molecules

$=2\times 10^{-3}\times 6.022\times 10^{23}=12.044\times 10^{20}$ molecules

Therefore, the correct option is A.

$500mL$ of a glucose solution contains

$6.02\times { 10 }^{ 22 }$ molecules. The concentration of the solution is

0%
$0.1M$
0%
$1.0M$
0%
$0.2M$
0%
$2.0M$

Explanation

Given,

$500ml$

$C_6H_{12}O_6$ contains

$6.02\times 10^{22}$ molecules

We know

$1$ mole of

$C_6H_{12}O_6$ contains

$6.02\times 10^{23}$ molecules

$\therefore$ No. of moles having

$6.02\times 10^{22}$ molecules=

$\cfrac {6.02\times 10^{22}}{6.02\times 10^{23}}=0.1$ moles

Now, Molarity=

$\cfrac {No.\quad of\quad moles}{V\quad in\quad L}=\cfrac{0.1 mol}{0.5L}=0.2M$

Concentration of solution=

$0.2M$ .

How many moles of electron weight one kilogram?

0%
$6.023\ \times \ { 10 }^{ 23 }$
0%
$\dfrac { 1 }{ 9.108 } \times \ { 10 }^{ 31 }$
0%
$\dfrac { 6.023 }{ 9.108 } \times \ { 10 }^{ 54 }$
0%
$\dfrac { 1 }{ 9.108\ \times \ 6.023 } \times \ { 10 }^{ 8 }$

Explanation

Mass of one electron

$\displaystyle =9.108 \times 10^{-31}$ kg.

1 mole of electrons

$\displaystyle = 6.023 \times 10^{23}$ electrons.

Mass of 1 mole of electrons

$\displaystyle =9.108 \times 10^{-31} \times 6.023 \times 10^{23}=9.108 \times 6.023 \times 10^{-8}$ kg.

$\displaystyle \dfrac {1}{9.108 \times 6.023 \times 10^{-8} }=\dfrac {1}{9.108 \times 6.023} \times 10^{8}$ electrons will weigh 1 kg.

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:

0%
$28\ g$
0%
$9.25\ g$
0%
$2.870\ g$
0%
$58\ g$

Explanation

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.

Sodium bicarbonate,

$NaHCO_{3}$ , can be purified by dissolving it in hot water

$(60^{\circ}C)$ , filtering to remove insoluble impurities, cooling to

$0^{\circ}C$ to precipitate solid

$NaHCO_{3}$ , and then filtering to remove the solid, leaving soluble impurities in solution. Any

$NaHCO_{3}$ that remains in the solution is not recovered. The solubility of

$NaHCO_{3}$ in hot water at

$60^{\circ}C$ is

$164\ g/litre$ and is

$69\ g/litre$ in cold water at

$0^{\circ}C$ . What is the percentage yield of

$NaHCO_{3}$ , when it is purified by this method?

0%
$55.34$ %
0%
$42.07$ %
0%
$69$ %
0%
$31$ %

Explanation

During crystallization

$5\%$ loss in acceptable expected yied

$=164-69=95\ g$

$5\%=\dfrac {x}{95}\times 100$

$x=\dfrac {5\times 95}{100}=4.75\ g$

actual yield will get

$=95-4.75=90.25\ g$

$\%\ NaHCO_3$ recverable

$=\dfrac {90.25}{164}\times 100$

$=55.03\%$ option

$(a)$

$H_3AO_4$ has 31.6% of "A" by mass. The atomic mass of the atom "A" is:

0%
17
0%
31
0%
35.5
0%
40

Explanation

Let the atomic mass of "A" be

$x$ .

Given,

31.6% of

$H_3AO_4=x$

$\Rightarrow \dfrac{31.6}{100}(1\times3+x+16\times4)=x$

$\Rightarrow 67+x=\dfrac{100}{31.6}x$

$\Rightarrow 67=\dfrac{68.4}{31.6}x$

$\Rightarrow x=31$

Hence, option

$B$ is correct.

What do carbon credits signify?

0%
Credits given in the course of carbon products sales
0%
Entitlements to emit certain quantity of green house gases
0%
Permissible amount of Carbon dioxide in the atmosphere
0%
The extent of carbon required to ensure sustainable development

The leaves of a green bean plant were exposed to carbon dioxide containing the radioactive isotope of carbon

$^{14}C$ for few hours in sunlight. A section of the plant's stem was then immediately obtained and placed on an X-ray film that blackens where exposed to radioactivity. The following diagrams show the cross-section of the stem, and the resulting autoradiograph. Why is structure

$P$ shown radioactive?

0%
$P$ transported radioactive carbohydrates from leaves to other plant parts
0%
Radioactive carbon dioxide gets dissolved in cell sap and is transported along with water and minerals by structure $P$
0%
The radioactive carbon is incorporated only in the tissues constituting structure $P$
0%
None of these

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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