MCQ Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 3

The molar mass of haemoglobin is about

$65000 \: g\:mol^{-1}$ . Every haemoglobin contains

$4$ iron atoms. Thus :

0%
iron content in haemoglobin is $0.35$ % by mass
0%
$1$ mole of haemoglobin contains $56$ g of iron
0%
$1$ mole of haemoglobin contains $224$ g of iron
0%
if Iron content is increased to $0.56$ %, molar mass of haemoglobin would be higher than $65000 \: g\: mol^{-1}$

Explanation

Iron in haemoglobin

$=\dfrac{4\times 56}{65000}\times 100=0.35$ %.
Iron in one mole haemoglobin

$= 4\times 56 = 224 \: g$ .
Molar mass of haemoglobin, if iron content is

$0.56$ %

$=\dfrac{100\times 4\times 56}{0.56}=40000<65000$
Thus, molar mass is decreased.

Which pair has the same percentage of carbon?

0%
$CH_3COOH \: and \: C_6H_{12}O_6$
0%
$CH_3COOH \: and \: C_{12}H_{22}O_{11}$
0%
$CH_3COOH \: and \: C_2H_5OH$
0%
$C_6H_{12}O_6 \: and \: C_{12}H_{22}O_{11}$

Explanation

The table below shows the simple ratio between the number of atoms.

$Compound$	$C$	$H$	$O$
$CH_3COOH$	1	2	1
$C_6H_12O_6$	1	2	1
$C_2H_5OH$	2	6	1
$C_{12}H_{22}O_{11}$	12	22	11

So,

$CH_3COOH$ and

$C_6H_{12}O_6$ have same % of

$C$ .

In diammonium hydrogen phosphate,

$(NH_4)_2HPO_4$ , percentage of :

0%
$P_2O_5$ is $53.78$ %
0%
$NH_3$ is $25.76$ %
0%
$P$ is maximum
0%
$N$ is maximum

Explanation

$1$ mole of

$(NH_4)_2HPO_4$ has

$0.5$ mole

$P_2O_5$ and

$2$ of

$NH_3$ .
Thus,

$1$ mole (

$132$ g)

$(NH_4)_2HPO_4$ has

$0.5$ mole (

$71$ g)

$P_2O_5$

$\%$ of

$P_2O_5=\frac{71}{132}\times 100=53.78$ %
Also,

$132$ g

$(NH_4)_2HPO_4$ has

$34$ g

$NH_3$ .
Thus,

$\%$ of

$NH_3=\dfrac{34}{132}\times 100=25.76$ %
Also,

$132$ g

$=28$ g

$N$ ,

$9$ g

$H$ and

$64$ g

$O$
Thus,

$\%$ of

$O$ is maximum.

$800$ g of a

$40\%$ solution by weight was cooled.

$100$ g of solute was precipitated. The percentage composition of remaining solution is :

0%
$31.4\%$
0%
$20.0\%$
0%
$23.0\%$
0%
$24\%$

Explanation

$\displaystyle Solute \: present \: in \: 800 \: g \: solution=800\times \frac{40}{100}$

$=320 \: g$

$Solute \: precipitated = 100 \: g$

$Solute \: left =220 \: g$

$\therefore Total \: weight \: of \: solution = 800 - 100 = 700 \: g$

$700 \: g \: solution \: has \: solute =220$

$\displaystyle 100 \: g \: solution \: has \: solute=\frac{220}{700}\times 100= 31.43$ %

The volume occupied by 7 grams of nitrogen gas at STP is:

0%
6.5
0%
5.6
0%
7.5
0%
4.5

Explanation

Number of moles of nitrogen

$= \dfrac{7}{28} = \dfrac{1}{4}$ moles

As we know, one mole occupies 22.4 litre volume.

Therefore,

$\dfrac{1}{4}$ mole will occupy,

$V = \dfrac{22.4}{4} = 5.6\ litre$

How many g of

$KCl$ would have to be dissolved in

$60$ g

${H}_{2}O$ to give

$20\%$ by mass of solution?

0%
$15$ g
0%
$1.5$ g
0%
$11.5$ g
0%
$31.5$ g

Explanation

As we know,

$\%$ by mass

$=\displaystyle\dfrac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$20 = \dfrac{\text{Mass of KCl} \times 100}{\text{Mass of KCl + Mass of water}}$

Let,

$x$ be the mass of KCl

$20=\dfrac{100x}{x+60}$

$x=15 g$

Therefore, mass of

$KCl$

$= 15$ g

Hence, the correct option is

$A$

Atomic number (Z) of a neutral atom and mass number (A) of an atom are equal to:

(Here n = number of neutrons and p = number of protons):

0%
Z = n and A = n+p
0%
Z = e and A = n+e
0%
Z = p and A = n+p
0%
Z = n and A = p+e

The average molar mass of air becomes more in presence of which gas if present in air :

0%
${H}_{2}$
0%
${N}_{2}$
0%
${C}_{2}{H}_{6}$
0%
$C{H}_{4}$

Explanation

Molar mass of air is

$28.80$ and among given gases, only

${C}_{2}{H}_{6}$ is heavier than air. So, it will increase the average molar mass of air.

Hence the correct option is C.

Find the weight of a substance if its molecular weight is

$70$ and in the gaseous form it occupies

$10$ L at

$27^oC$ and

$760$ mm

$Hg$ pressure.

0%
32.5 g
0%
35.5 g
0%
28.38 g
0%
None of the above

Explanation

We know that 760 mm of hg is equal to 1 atm, so we use 1 atm.

As we know,

$Pv=nRT\\ 1\times 10=\frac { w }{ 70 } \times 0.082\times 300\\ w=28.38g$

Which facts are revealed from Avogadro's law for ideal gases?

0%
The distance between molecules is much large than the actual dimension of molecules.
0%
Equal number of molecules of different gases under identical conditions of temperatures and pressure occupy the same volume.
0%
Saturated vapours obey gas laws
0%
1 mole of a gas contains $6.023\times 10^{23}$ molecules of gases.

Explanation

Avogadro's Law :
It states "equal volumes of any two gases at the same temperature and pressure contain the same number of molecules".

$V \propto n$ (At constant P and T)

$\dfrac{V_1}{V_2} = \dfrac{n_1}{n_2}$

and each gas's one mole contains

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

The percentage of

${P}_{2}{O}_{3}$ in

${H}_{3}P{O}_{3}$ is:

0%
$44.69$
0%
$33.52$
0%
$67.07$
0%
$50.04$

Explanation

The reaction involved is as follows:

${P}_{2}{O}_{3} + 3{H}_{2}O\rightarrow 2{H}_{3}P{O}_{3}$
So, two moles of

$H_3PO_3$ has one mole of

$P_2O_3$ .

$\therefore \%$ of

${P}_{2}{O}_{3}\ in\ {H}_{3}P{O}_{3} = \displaystyle\frac{110\times 100}{164} =$

$67.07\%$

The total ionic strength of the solution containing

$0.1$ M of

$CuS{O}_{4}$ and

$0.1$ M of

${Al}_{2}{(S{O}_{4})}_{3}$ is :

0%
0.2 M
0%
0.19 M
0%
0.8 M
0%
0.105 M

Explanation

In solution,

$CuSO_4\rightarrow Cu^{+2} + SO_4^{-2}$
0.1 0.1 0.1

$Al_2(SO_4)_3 \rightarrow 2Al^{+3} + 3SO_4^{-2}$
0.1 0.2 0.3
Total ionic strength =

$\displaystyle\frac{1}{2}\Sigma c{Z}^{2}$
=

$\displaystyle\frac{1}{2}[0.1\times {2}^{2} +0.1\times {2}^{2}+ 0.1\times 2\times {3}^{2}+ 0.1\times 3\times {2}^{2}] = 0.19$

In diammonium phosphate

$(NH_4)_2HPO_4$ , the percentage of

$P_2O_5$ is :

0%
35.87
0%
46.44
0%
51.99
0%
53.78

Explanation

Molar Mass of Diammonium Phosphate

$(NH_4)_2 HPO_4$ is

$28+8+1+31+64=132 \ grams$

Molar Mass of

$P_2O_5$ is

$62+80=142 \ grams$

We know that

$2$ moles of diammonium hydrogen sulphate yields

$1$ moles of

$P_2O_5$ i.e

$2(NH_4)_2HPO_4 \rightarrow P_2O_5$

Mass of

$2$ moles of

$P_2O_5=2 \times 132 \ grams = 264 \ grams$

Therefore, percentage composition of

$P_2O_5 = \cfrac {Mass \ of \ P_2O_5}{Mass \ of \ 2 \ moles \ of \ (NH_4)_2HPO_4}$

$=\cfrac {142}{264} \times 100 = 53.78$ %

Calculate the volume occupied by 40 g. of a

$CH_4$ at S.T.P if its V.D. is 8.

0%
24 litres
0%
46 litres
0%
34 litres
0%
56 litres

Explanation

Molar mass

$=2\times vapour\ density = 2\times8 = 16 gm$
No of moles

$=\frac{40}{16} = 2.5$
So volume

$= 2.5\times22.4 = 56\ litres$

The volume occupied by 53.5 g of

$Cl_2$ is :

0%
15.76 litres
0%
16.87 litres
0%
18.65 litres
0%
20 litres

Explanation

One mole of a gaseous substance occupies

$22.4 \ L$ sovolume occupied by

$53.5$ g of

$Cl_2 = 53.5\times\dfrac{22.4}{71} = 16.87 \ L$ .

$IE_{1}$ for

$He$ atom is 24.6 eV. The energy required to remove both the electrons from an atom in its ground state will be:

0%
59 eV
0%
81 eV
0%
79 eV
0%
None of these

Elements

$X, Y$ and

$Z$ have atomic numbers 5, 9 and 11 respectively. Which one forms an anion?

0%
$X$
0%
$Y$
0%
$Z$
0%
Both $B$ and $C$

Explanation

Atomic number of

$X$ is

$5$ and electronic configuration is:

$2, 3$

Atomic number of

$Y$ is

$9$ and electronic configuration is:

$2, 7$

Atomic number of

$Z$ is

$12$ and electronic configuration is:

$2, 8, 1$

$Y$ is one electron short of stable completely filled configuration and thus readily accepts one electron to form an anion.

A vessel contains N molecules of oxygen at a certain temperature and pressure. How many molecules of sulphur dioxide can the vessel accommodate at the same temperature and pressure.

0%
N
0%
2N
0%
N/2
0%
4N

Calculate the volume occupied by 0.01 moles of helium gas at STP.

0%
2.24 L
0%
0.224 L
0%
22.4 L
0%
None of the above

Explanation

As we know, one mole of a gaseous substance occupies 22.4 litre, so volume occupied by 0.01 mole helium is

$0.224 \ L$ .

The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditions of temperature and pressure. If 20 litres of nitrogen contains 'x' no. of molecules, state the no. of molecules in 20 litres of ammonia.

0%
x
0%
x/2
0%
x/4
0%
2x

What is the volume occupied by X molecules of

$N_2$ ?

0%
2V
0%
4V
0%
V/2
0%
V

Explanation

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

$N_2$ occupies V litre volume.

An experiment was set up to determine the percentage of water absorbed by raisins. If the mass of dry raisins was

$40 g$ and mass of wet raisins was

$45 g$ , the percent water absorbed would be:

0%
$\displaystyle\frac{45}{40}\times 100$
0%
$\displaystyle\frac{40}{45}\times 100$
0%
$\displaystyle\frac{45 - 40}{40}\times 100$
0%
$\displaystyle\frac{45 - 40}{45}\times 100$

Explanation

${W}_{1} =$ initial weight of raisins,

$40 g$

${W}_{2} =$ final weight of wet raisins,

$45 g$

$\displaystyle\frac{{W}_{2} - {W}_{1}}{{W}_{1}}\times 100$ .

$\displaystyle\frac{{45} - {40}}{{45}}\times 100$

A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. Under the same conditions of temperature and pressure, how many molecules of nitrogen gas would be present in the same vessel?

0%
2X
0%
X
0%
0.5X
0%
3X

Glucose is a physiological sugar. If the mass percentage of

$C=x\%$ , mass percent of

$H=y\%$ and mass percent of

$O=z\%$ in glucose

$(C_6H_{12}O_6)$ , then the value of

$x,\ y$ and

$z$ are, respectively:

0%
$40,\ 6.67,\ 53.33$
0%
$30,\ 6.67,\ 43.33$
0%
$40,\ 7.67,\ 63.33$
0%
none of the above

Explanation

Molecular weight of glucose

$=C_6H_{12}O_6=12\times 6+1\times 12+16\times 6$

$=180$ g

$180$ g of glucose contains

$72$ g of

$C$ .
Percnetage of

$C=\dfrac {72\times 100}{180}=40\%$

Percentage of

$H=\dfrac {12\times 100}{180}=6.67\%$

Percentage of

$O=\dfrac {96\times 100}{180}=53.33\%$

Insulin contains

$3.4$ % sulphur. The minimum molecular mass of insulin is :

0%
$940$
0%
$350$
0%
$470$
0%
$560$

Explanation

Correct answer: Option $A$

Explanation:

Since an atom is the smallest discrete unit of nature, a molecule of insulin must have at least one

$S$ atom, it cannot have half an

$S$ atom. Therefore, the minimum molecular mass of insulin will be when it has exactly one

$S$ atom, which weights

$32g$

$\implies 3.4\% = 32g$ of

$S$

$\implies 100\% = \dfrac{32}{3.4} \times 100\%$

$\implies 100\% = 941.17g$

So Minimum molar mass of insulin is around

$940 g/mol$

Hence, option $A, 940$ is correct.

$100$ mL of

$H_2SO_4$ and

$100$ mL of

$H_2O$ are mixed, the mass percent of

$H_2SO_4$ in the resulting solution will be:

$(d_{H_2SO_4}=0.09\: g\: mL^{-1}, d_{H_2O}=1.0\: g\: mL^{-1})$

0%
$90.32$ %
0%
$47.36$ %
0%
$50.56$ %
0%
$60$ %

Explanation

Mass of

${H_2SO_4}=\dfrac {100\times 0.9}{100\times 0.9+100\times 1}=\dfrac {90}{190}=0.4736$ g.
% Mass of

$H_2SO_4=43.36$ %

The molar mass of hemoglobin is about

$65000 \: gmol^{-1}$ . Every hemoglobin contains

$4$ iron atoms. Thus:

0%
iron content in haemoglobin is $0.35$ %.by mass
0%
$1$ mole of haemoglobin contains $56$ g iron
0%
$1$ mole of haemoglobin contains $224$ g iron
0%
if iron content is increased to $0.56$ %, molar mass of hemoglobin would be higher than $65000 \: g\: mol^{-1}$

Explanation

Iron in hemoglobin

$=\dfrac{4\times 56}{65000}\times 100=0.35$
Iron in one mole hemoglobin

$= 4\times56 = 224 \: g$
Molar mass of haemoglobin if iron content is

$0.56$ %

$\displaystyle =\frac{100\times 4\times 56}{0.56}=40000<65000$
Thus, molar mass is decreased.

The molality of 1 L solution with x%

$H_2SO_4$ is equal toThe weight of the solvent present in the solution is 910 g. The value of x in g per 100 mL is :

0%
90
0%
80.3
0%
40.13
0%
9

Explanation

$Molality=\dfrac {100\times \text {Weight of solute}}{Mw\times \text {Weight of solvent}}$

$9=\dfrac {1000\times W}{98\times 910}$

$W=802.6\ g\ L^{-1}$

$=80.26\ g\ per\ 100\ mL$

$13.4$ g of a sample of unstable hydrated salt,

$Na_2SO_4.xH_2O$ was found to contain

$6.3$ g of

$H_2O$ . The number of molecules of water of crystallisation is :

0%
$5$
0%
$7$
0%
$2$
0%
$10$

Explanation

Weight of salt

$=13.4$ g
Weight of

$H_2O=6.3$ g
Weight of anhydrous salt

$=13.4-6.3=7.1$ g
Moles of anhydrous salt

$=\dfrac {7.1}{842}=0.05$
Moles of

$H_2O=\dfrac {6.3}{18}=0.35$ mol

$0.05$ mol of anhydrous salt gives

$0.35$ mol of

$H_2O$ .

$1$ mol of anhydrous salt will give

$\dfrac {0.35}{0.05}=7$ mol of water.

"Suvarnabhasm", an ayurvedic drug, is found to contain

$400$ ppm of colloidal gold. Mass % of gold (atomic mass of Au

$=197$ ) will be :

0%
$0.040\: %$
0%
$7.88\: %$
0%
$0.0788\: %$
0%
$4\times 10^{-4}$ %

Explanation

Mass % of gold means amount of gold present in

$100$ g.
As it is given that, in

$10^6$ g drug, gold present is

$400$ g, so in

$100$ g drug, gold is

$\dfrac{400\times 100}{10^6} = 0.040$ % of the compound.

Four 1-1 litre flasks are separately filled with the gases

$H_2,\, He,\, O_2$ and ozone. The ratio of total number of atoms of these gases present in different flask would be :

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

Explanation

According to Avogadro's law, all gases at same temperature and pressure will contain same number of molecules. Hence, number of molecules of all gases will be same bt ration of atoms of each gases will

$2:1:2:3$ .

A sample of

$H_{2}SO_{4}$ contains

$3.2$ kg of sulphur. The weight (in g) of hydrogen present in the sample is:

0%
100
0%
200
0%
50
0%
150

Explanation

$H_{2}SO_{4}$ no of moles of hydrogen=2

$\times$ no of moles of hydrogen. Now 3.2kg of sulphur=3200 g of sulphur=100 moles of sulphur.

So, no fo moles of hydrogen=200 moles. So weight of the hydrogen=(1

$\times$ 200)=200 gm. So the correct option is B

Statement: Equal volumes of all gases always contain equal number of moles.

State whether the given statement is true or false.

0%
True
0%
False

Explanation

Avogadro law: It states that equal volumes of all gases under the same conditions of temperature and pressure contain the equal number of molecules.

This means that as long as the temperature and pressure remain constant, the volume depends upon number of molecules of the gas.

Since volume of a gas is directly proportional to the number of moles; one mole of each gas at standard temperature and pressure (STP) will have same volume.

Hence, the given statement is

$\text{true}$

State True or False.

Mass of reactants is always equal to the mass of the products. (Don't consider nuclear reaction)

0%
True
0%
False

Explanation

The statement, "Mass of reactants is always equal to the mass of the products. (Don't consider nuclear reaction)" is true. This is the statement of the law of conservation of mass.
For example, consider the combustion of

$C$ .

$C+O_2 \rightarrow CO_2$
The molar masses of

$C$ ,

$O_2$ and

$CO_2$ are 12 g/mol, 16 g/mol and 44 g/mol respectively.
Total mass of reactants for combustion of 1 mole of

$C$ is

$12+32=44 \: g$ .
The total mass of products for combustion of 1 mole of

$C$ is

$44 \: g$ .

Hence the mass of the reactants is equal to the mass of the products.

Law of conservation of mass was put forward by :

0%
Lavoisier
0%
Dalton
0%
Priestly
0%
Thomson

Two elements always combine together in a fixed ratio by weight.

0%
True
0%
False

Explanation

The statement, "Two elements always combine in a fixed ratio by weight." is false. In a compound, their ratio of masses can remain fixed. But two elements can combine in different ratios to form different compounds.
For example, consider C and O. They combine to form

$CO$ and

$CO_2$ .
12 g of C combine with 16 g of O to form

$CO$ . The ratio by weight of C to O is

$12:16$ or

$3:4$ .
12 g of C combine with 32 g of O to form

$CO_2$ . The ratio by weight of C to O is

$12:32$ or

$3:8$ .
Thus, elements can combine in different ratios to form different compounds but the ratio remains fixed for a particular compound.

State True or False.

Equal volumes of different gases under similar conditions of temperature and pressure contain equal number of atoms.

0%
True
0%
False

Atomic mass of an element is:

0%
actual mass of one atom of the element
0%
average mass of an atom of different atoms of the element
0%
always a whole number
0%
None of these

Under the same conditions, two gases have the same number of molecules. They must:

0%
be noble gases
0%
have equal volumes
0%
have a volume of 22.4 dm $^3$ each
0%
have an equal number of atoms

$1$ L of

$O_2$ at

$15^oC$ and

$750$ mm pressure contains

$\text N$ molecules, the number of molecules in

$2$ L of

$SO_2$ under the same conditions of temperature and pressure will be :

0%
$\dfrac {\text N}{2}$
0%
$\text N$
0%
$2\text N$
0%
$4\text N$

Explanation

Under similar conditions of temperature and pressure, equal volume of gas contains equal number of molecules.

$\therefore 1\ L=N$ molecules

$2\ L=2\ N$ molecules

Which of the following is the best thing to do during heavy lighting?

0%
Lying on the ground in an open place
0%
Going into the nearest water body
0%
Staying indoors away from metallic doors or windows
0%
Standing under a tall tree

The percentage value of nitrogen in urea is about :

0%
$46$
0%
$85$
0%
$18$
0%
$28$

Explanation

The formula of urea is

$\displaystyle NH_{2}CONH_{2}$

$\displaystyle W_{N}=28$

$\displaystyle W_{C}=12$

$\displaystyle W_{O}=16$

$\displaystyle W_{H}=4$
In

$60$ g urea,

$\displaystyle W_{N}=28g$

$100$ g urea,

$\displaystyle W_{N}$ is =

$\displaystyle \frac{28}{60}\times 100 = 46\%$

3 g of a hydrocarbon on combustion in excess of oxygen produces 8.8 g of

$CO_2$ and 5.4 g of

$H_2O$ .

The data illustrates the law of :

0%
Conservation of mass
0%
Multiple proportions
0%
Constant proportions
0%
None of the above

Explanation

Mass of carbon in 8.8 g

$CO_2 = \dfrac {12}{44} \times 8.8 = 2.4 \ g$

Mass of hydrogen in 5.4 g

$H_2O = \dfrac{2}{18} \times5.4 = 0.6 \ g$

Total mass

$( C + H )= 2.4 + 0.6 = 3.0 \ g$

Mass of hydrocarbon = 3.0 g (given)

These data illustrate the law of conservation of mass.

Option (A) is correct.

The weights of two elements which combines with one another are in the ratio of their:

0%
atomic weight
0%
molecular weight
0%
equivalent weight
0%
none of the above

Explanation

According to the law of multiple proportions, when two elements combine to form more than one compound, the weight of one element combined with a fixed weight of the other is in a ratio of small whole numbers.

Hence, the weight of two elements that combine is in the ratio of their atomic weight.

So, the correct option is

$A$ .

Modern atomic mass scale is based on the mass of:

0%
$H - 1$
0%
$C - 12$
0%
$C - 14$
0%
$C - 16$

Explanation

Modern atomic weight scale is based on

$C - 12$ .
The standard unit for expressing the mass of atom is amu (atomic mass unit).
It is equal to 1/12 of the mass of an atom of carbon-12 equal to

$1.6605 \times 10^{-19} \: g$ .
amu is also called as avogram.
Avogram is a unit of mass and weight equal to one gram divided by the Avogadro's number.

Insulin contains

$3.4$ % sulphur. The minimum molecular weight of insulin is :

0%
941.176
0%
944
0%
945.27
0%
none

Explanation

$3.4$ g

$S$ present in

$100$ g insulin.

$\therefore 32$ g

$S$ is present in

$\displaystyle \frac{100}{3.4}\times 32 = 941.176$ , which is equal to the minimum molecular weight of insulin.

Photo-electric cell is not used in:

0%
television
0%
photography
0%
reproduction of sound in cinema
0%
automatic switching of street lightening circuits

Explanation

$\text{The exposure meter instead of photo-electric cell is used in camera to know the correct time of exposure}$

Percentage of Se in peroxidase anhydrous enzyme is

$0.5$ % by weight (At. wt. =

$78.4$ ) then, minimum molecular weight of peroxidase anhydrous enzymes is :

0%
$\displaystyle 1.568\times 10^{4}$
0%
$\displaystyle 1.568\times 10^{3}$
0%
$15.68$
0%
$\displaystyle 2.136\times 10^{4}$

Explanation

In peroxidase anhydrous enzyme 0.5% Se is present means, 0.5 g Se is present in 100 g of enzyme. In a molecule of enzyme one Se atom must be present. Hence 78.4 g Se will be present :

$0.5$ g Se present in

$100$ g enzyme.

$\therefore 78.4$ g present in

$\displaystyle \frac{100}{0.5}\times 78.4 =$ Molecular weight

The density of a salt solution is

$1.13 g cm^{-3}$ and it contains

$18\%$ of

$NaCI$ by weight The volume of the solution containing

$36.0\ g$ of the salt will be

0%
$200 cm^3$
0%
$217 cm^3$
0%
$177 cm^3$
0%
$157 cm^3$

Explanation

Consider the volume of the solution

$=x cm^3$
Then the mass of the solution will be

$= 1.13x$

$(mass = density\times volume)$
The solution contains 18% of NaCl by weight

$\therefore \frac {18}{100}\times 1.13x=36$

$x=\frac {3600}{18\times 1.13}=177 cm^3$

$10\ g$ of a crystalline metal sulphate salt when heated generates approximately

$6.4\ g$ of an anhydrous salt of the same metal. The molecular weight of the anhydrous salt is

$160\ g$ . The number of water molecules present in the crystal is

0%
$1$
0%
$2$
0%
$3$
0%
$5$

Explanation

Consider that the salt contains

$\times$ molecules of water.
Molecular weight of anhydrous salt

$= 160 g$
so molecular weight of hydrated salt will be

$=160+18xg$
Then, no. of moles of water present in 10x gm of hydrated salt

$=\frac {10}{160+18x}\times x$
and weight of water present in 10 gm of hydrated salt

$=\frac {10x}{160+18x}\times 18$

$\underset {10 g}{Hydrated} salt\rightarrow \underset {604 g}{Anhydrous} salt+\underset {3.6 g}{Water}$

$\frac {180x}{160+18x}=3.6$

$180x=576+64.8 x$

$x=5$

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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