MCQ Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 10

$5.25%$ solution of a substance is isotonic with a

$1.5\%$ solution of urea (molar mass =

$60g mol^{-1})$ in the same solvent. If the densities of both the solutions are assumed to be equal to

$1.0 g cm^{-3}$ , molar mass of the substance will be:

0%
$105.0 g mol^{-1}$
0%
$210.0 g mol^{-1}$
0%
$90.0 g mol^{-1}$
0%
$15.0 g mol^{-1}$

Explanation

Since the 2 solutions are isotonic :

$\Rightarrow C_1=C_2$

$\Rightarrow \dfrac{5-25}{M}=\dfrac{1.5}{60}$ [

$\therefore$ Densities are equal, volume can be ignored ]

$\therefore M=210g/mol.$

Hence, the answer is

$210g/mol.$

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

[n and m are the masses of reactants and p and q are the masses of products formed.]

0%
$n-m=p-q$
0%
$n+m=p+q$
0%
$n=m$
0%
$p=q$

Explanation

Law of conservation of mass:- Mass is neither created nor destroyed it just from one compound to other.

$n+m\rightarrow$

$p+q$

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

Option

$B$ is correct.

$27\ gm\ Al$ is heated with

$49\ ml$ of

$H_{2}SO_{4}$ (specific gravity

$=2$ ) produces

$H_{2}$ gas. Percentage of

$Al$ reacted with

$H_{2}SO_{4}$ is:

0%
$50\%$
0%
$66.66\%$
0%
$75\%$
0%
$80\%$

Explanation

$2Al+3H_2SO_4\longrightarrow Al_2{(SO_4)}_3+3H_2\uparrow$

$2$ moles of

$Al$ reacts with

$3$ moles of

$H_2SO_4$ .

i.e.

$2\times 27g$ reacts with

$3\times 98g$

$H_2SO_4$ .

Mass of

$H_2SO_4$ present

$=49\times 2=98g$

$\therefore$ Mass of

$Al$ reacted

$=\cfrac{2}{3}\times 27=18g$

$\therefore \%$

$Al$ reacted

$=\cfrac{18}{27}\times 100=66.66\%$

The amount of water obtained from combustion of

$16 grams$ of methane is_______

0%
16g
0%
36g
0%
32g
0%
62g

Explanation

$CH_4(g)+2O_2\rightarrow CO_2(g)+2H_2O(g)$

According to the above equation, one mole of

$CH_4(16\ g)$ on combustion gives one mole of carbon dioxide and 2 moles of water and 2 moles of water

$=2\times(2+16)=36\ g$ of water.

Our body contains ___________ of water by weight.

0%
$70\%$
0%
$60\%$
0%
$50\%$
0%
$80\%$

Explanation

Our body contains

$\underline{70\%}$ of water by weights.

$H_2SO_4$ is

$98\%$ by weight of solution .Hence it is ?

0%
1 molal
0%
10 molal
0%
50 molal
0%
500 molal

Explanation

Let the weight of solution be

$100g$

Then the weight of

$H_2SO_4$ will be

$98g$

$\therefore$ Number of moles of

$98$ %

$H_2SO_4$

$=\cfrac {\text{Weight}}{\text{Molecular mass}}$

$=\cfrac {98}{98}=1$

Molality=

$\cfrac {\text{No. of moles of solute}}{\text{Weight of solvent}}$ (Weight of

$H_2O=100-98=2g$ )

$=\cfrac {1}{2}$

$=\cfrac {1}{2}\times 1000=500$ molal solution (Converting into kg)

If mass percentage of

$Mg^{2+}$ in a biomolecule is 0.2%, then the number of

$Mg^{2+}$ ions present in 10 g of biomolecule is (Atomic weight of Mg = 24)

0%
$\frac{1}{2} \times 10^{-2}N_{A}$
0%
$\frac{1}{12} \times 10^{-2}N_{A}$
0%
$\frac{1}{4}N_{A}$
0%
$\frac{1}{12}N_{A}$

Explanation

$100$ % of biomolecule contains-

$0.2$ % of

$Mg^{2+}$

So,

$10g$ of biomolecule contains-

$\cfrac {2}{100\times 10}\times 10$

$\Rightarrow2\times 10^{-2}gMg^{2+}$

$10g$ of biomolecule contains-

$2\times 10^{-2}gMg^{2+}$

$\Rightarrow24g\longrightarrow N_A$

$\Rightarrow2\times 10^{-2}g\longrightarrow x$

$\Rightarrow x\longrightarrow \cfrac {2\times 10^{-2}g\times N_A}{24}$

$\Rightarrow x\longrightarrow \cfrac {1}{12}\times 10^{-2}N_A$

The number of

$Mg^{2+}$ ions present in

$10g$ of biomolecules is

$\cfrac {1}{12}\times 10^{-2}N_A$ .

What is the proportion by weight the constitution elements of

$Cu$ and

$O$ in the compound

$CuO$ ?

0%
$3.971: 1$
0%
$4 : 2.968$
0%
$4 : 3.971$
0%
$2.968 : 2$

Explanation

Atomic mass of

$Cu=63.546\mu$

Atomic mass of

$O=16\mu$

$CuO$ , the proportion by weight is,

$\dfrac { 63.546 }{ 16 } :\dfrac { 16 }{ 16 } \equiv 3.971:1$

Suppose 30% molecules have M = 20000; 40% molecules have M= 30000, rest of them have M=then the PDI is :

0%
1.45
0%
1.20
0%
0.83
0%
0.98

How many grams of concentrated sulphuric acid

$(80\%W/W)$ sould be used to prepare 100 mL of one molar

$H_{2}SO_{4}$ solution?

0%
10 g
0%
12. 25g
0%
15 g
0%
20 g

200 gram of a solution of water and sugar contains 20 g solute. What is the concentration of the solution in mass percentage?

0%
20%
0%
10%
0%
40%
0%
5%

Explanation

Concentration of Solute =

$\dfrac{20}{200} \times 100$ = 10 %

If mass percentage of iron in a biomolecule is 0.4% then the minimum molecular weight of biomolecule will be (Atomic weight of Fe = 56)

0%
18000
0%
16000
0%
10000
0%
14000

$40$ % of a mixture of

$0.2mol$ of

${N}_{2}$ and

$0.6mol$ of

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

${NH}_{3}$ according to the equation:

${N}_{2}(g)+3{H}_{2}(g)\rightleftharpoons 2{NH}_{3}(g)$
at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are

0%
$4:5$
0%
$5:4$
0%
$7:10$
0%
$8:5$

The actual molecular mass of chlorine is :

0%
$58.93 \times {10^{ - 24}}g$
0%
$117.86 \times {10^{ - 24}}g$
0%
$58.93 \times {10^{ - 24}}kg$
0%
$117.86 \times {10^{ - 24}}kg$

Explanation

$1$ Atom of chlorine using

$=35.453$ amu

$1$ Molecule of chlorine has formula

$C12$

$\therefore$ Molecular mass of

$Cl_2 = 2\times 35.453$ amu

$=70.906$ amu

$1$ amu

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

so,

$70.906\ \times 1.66\times 10^{-24}\ g$

$=177.7\times 10^{-24}g$

The weight of

$C{H_4}$ in 9-L cylinder at

${27^0}C$ temperature and

$16$ atm pressure is (

$R=0.08$ L atm

${K^{ - 1}}\,mo{l^{ - 1}}$ ).

0%
$9.6$ g
0%
$96.0$ g
0%
$4.8$ g
0%
$48.0$ g

Explanation

Using the ideal gas equation,

$PV=nRT$

$16atm\times 9L=n\times 0.082\times 300K$ .....(given)

Therefore, n=6

$n=\dfrac{Weight}{atomic mass}$

weight of

$CH_4=16g{mol}^{-1}\times 6mol=96g$

$0.60g$ sample consisting of only

$Ca{C}_{2}{O}_{4}$ and

$Mg{C}_{2}{O}_{4}$ is heated at

${500}^{o}C$ , converting the two salts of

$Ca{CO}_{3}$ and

$Mg{CO}_{3}$ . The sample then weighs

$0.465g$ . If the sample had been heated to

${900}^{o}C$ , where the products are

$CaO$ and

$MgO$ , what would the mixtures of oxides have weighed?

0%
$0.12g$
0%
$0.21g$
0%
$0.252g$
0%
$0.3g$

Zinc sulphate contains 22.65% Zn and 43.9%

$\mathrm { H } _ { 2 } \mathrm { O }$ . if the law of constant proportions is true, then the mass of Zinc required to give 40g crystal will be

0%
90.6 g
0%
9.06 g
0%
0.906 g
0%
906 g

60 g of solution containing 40% by mass of

$NaCl$ are mixed with 100 g of a solution containing 15% by mass

$NaCl$ . Determine the mass percent of sodium chloride in the final solution.

0%
24.4 %
0%
78 %
0%
48.8 %
0%
19.68 %

Explanation

$Mass \ of \ NaCl \ in \ 40\% \ solution=\dfrac{40}{100}\times{60}=24g$

$Mass \ of \ NaCl \ in \ 15\% \ solution=\dfrac{15}{100}\times{100}=15g$

$Total \ mass \ of \ NaCl \ in \ resulting \ solution=24+15=39g$

$Total \ mass \ of \ final \ solution=60+100=160g$

$Mass \ percent \ of \ NaCl \ in \ final \ solution=\dfrac{39}{160}\times{100}=24.4\%$

The number of

$H_{3}O^{+}$ ions present in 10ml of water at

$25^{\circ}C$ is:

0%
$6.023 \times 10^{-14}$
0%
$6.023 \times 10^{14}$
0%
$6.023 \times 10^{-19}$
0%
$6.023 \times 10^{19}$

Explanation

No. of hydronium ions present in

$10$ ml water at

${ 25 }^{ 0 }C$ .

$pH=7$ at

${ 25 }^{ 0 }C$

$\left[ { H }^{ + } \right] ={ 10 }^{ -7 }M$

No. of moles

$=$ Concentration

$\times$ Volume

$=\left( { 10 }^{ -7 } \right) \left( 10 \right) \times { 10 }^{ -3 }$

$={ 10 }^{ -9 }$ moles

1 mole has

$6.023\times { 10 }^{ 23 }$ ions of hydronium.

$\Rightarrow \quad 6.023\times { 10 }^{ 23 }\times { 10 }^{ -9 }$ ions in

$10$ ml water

$\Rightarrow 6.023\times { 10 }^{ 14 }$ ions are there in

$10$ ml water at

${ 25 }^{ 0 }C$ .

Hence, option B is correct.

Which of the following represents the Avogadro number?

0%
Number of molecule present in 1 L of gas at N.T.P
0%
Number of molecule present in 22.4 L of gas at N.T.P
0%
Number of molecule present in 22.4 of gas at 298K and 1 atm pressure
0%
Number of molecule present in one mole of gas at any temp and pressure

Explanation

Avogadro number is defined as the number of gas molecules present in one mole of gas at any pressure and temperature. It's value is

$6.023 \times 10^{23}$ . So the correct option is D

What is the value of

$n$ in the following equation?

$Cr\left( OH \right) _{ 4 }^{ - }+OH^{ - }\longrightarrow Cr{ O }_{ 4 }^{ 2- }+H_{ 2 }O\ +\ ne^-$

0%
$3$
0%
$6$
0%
$5$
0%
$2$

Explanation

The given reaction is

$[Cr(OH)_{4}]^{-}+ OH^{-} \rightarrow [CrO_{4}]^{-2} +H_{2}O+ne^{-}$

Now balancing the equation,

$2[Cr(OH)_{4}]^{-}+ 8OH^{-} \rightarrow 2[CrO_{4}]^{-2} +8H_{2}O+6e^{-}$ . So the value of n is 6. Correct option is B

A sample of potato starch was ground in ball mill to give a starch lime molecule of lower molecular weight. The product analysed

$0.86\%$ phosphorus. If each molecule is assumed to contain one atom of phosphorus.The molecular weight of the material is:-

0%
$3.6\times 10^{6}$
0%
$4.9\times 10^{4}$
0%
$7.2\times 10^{6}$
0%
$3.6\times 10^{4}$

The percentage loss in weight after heating a pure sample of potassium chlorate (mol.

$wt.=122.5$ ) will be:

0%
12.25
0%
24.50
0%
39.18
0%
19.6

Explanation

$2KClO_3\longrightarrow 2KCl+3O_2$

$2\times 122.5$

$g$

$3\times 32$

$g$

$O_2$

$\implies 2\times 122.5$

$g$

$KClO_3$ show loss of

$3\times 32$

$g$ of

$O_2$

$100$

$g$

$KClO_3$ shows weight loss of

$\cfrac {3\times 32\times 100}{2\times 122.5}=39.17$

$\implies \%$ Loss of weight

$=39.17$

A solution containing

$500 g$ of a protein per litre is isotonic with a solution containing

$3.42 g$ of sucrose per litre. The molecular mass of protein is:

0%
$5$
0%
$146$
0%
$34200$
0%
$50000$

Explanation

Two solutions are said to isotonic when they are at the same concentrations.

The molar mass of sucrose

$=342$

$C_1 =$ Concentration of protein

$C_2 =$ Concentration of sucrose

Since, they are isotonic,

$C_1 = C_2$

$\cfrac {500}{\text{Molar mass of protein}} = \cfrac {3.42}{342}$

$\implies$ Molar mass of protein

$=500\times 100=50000$

$g$

The carbonate of a metal is isomorphous with

$MgCO_{3}$ and contains

$6.091\%$ of carbon. Atomic weight of the metal is nearly:

0%
$48$
0%
$68.5$
0%
$137$
0%
$120$

Explanation

The molecular formula of metal carbonate is ${MCO}_{3}$ as it is isomorphous with ${MgCO}_{3}$ .

Let the mass of metal be $M$ . Molecular weight of this carbonate $= (M+12+48)$ . Mass percentage of carbon in this compound $= 6.091%$

$\dfrac{6.091}{100}\times(M+12+48) = 12$

$M = \dfrac{12\times100}{6.091} - 60$

$M = 137$

Atomic wt. of the element is 137.01 amu which is Barium and the metal carbonate is BaCO $_3$

Hence, the correct option is $C$

A compound contains

$28\%$ Nitrogen, by mass. The possible molecular mass of the compounds is/are?

0%
$100$
0%
$125$
0%
$150$
0%
$140$

Which of the following weights the most?

0%
$10^{23}$ molecules of $H_{2}$
0%
$1$ mole of $H_{2}O$
0%
$1$ mole of $N_{2}$
0%
$10^{22}$ atoms of oxygen

Hemoglobin contains 0.334 % of iron by weight. The molecular weight of hemoglobin is approximately 67200.The number of iron atoms (Atomic weight of Fe is 56) present in one molecule of hemoglobin is

0%
4
0%
6
0%
3
0%
2

Effective atomic number of the central metal atom in the compound/complex

$\mathrm { Cr } ( \mathrm { CO } ) _ { 6 }$ is:

0%
34
0%
35
0%
36
0%
37

If an iodized salt contains

$1\%$ of

$KI$ and a person takes

$2\ g$ of the salt every day, the iodine ions going into his body every day would be approximately

$(K=39, I=127)$

0%
$9.5 \times 10^{21}$
0%
$7.2 \times 10^{19}$
0%
$7.2 \times 10^{21}$
0%
$9.5 \times 10^{19}$

Explanation

Mass of

$KI=2\times \dfrac {1}{100}$

$=0.02\ gm$

$KI\to K^+ +I^-$

No. of iodine ion

$=1$

$\therefore$ Total no of

$I^-$ ion

$=\dfrac {0.02}{127+39}\times 6.02\times 10^{23} \times 1$

$=\dfrac {0.02}{16}\times 6.02\times 10^{23}$

$7.2\times 10^{19}$ ions

option

$(b)$

1 atomic mass unit is equal to:

0%
$\tfrac{1}{12}^{th}$ mass of a carbon-12 atom
0%
$1.66\times 10^{-24}\ g$
0%
$6.023\times 10^{-23}\ g$
0%
$6.023\times 10^{23}\ g$

Explanation

One atomic mass unit is a mass unit equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12.

So, we can write 1 atomic mass unit

$= \dfrac {1}{12}$ mass of carbon-12 atom

$=\dfrac {1}{12}$

$\times$

$\dfrac {12}{{6.023 \times 10^{23}}} g$

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

The critical density of the gas

$CO_2$ is 0.44

$g\ cm^{–3}$ at a certain temperature. If

$r$ is the radius of the molecule,

$r^3$ in

$cm^3$ is approximate:

[N is Avogadro number]

0%
$\cfrac {25} {\pi N}$
0%
$\cfrac {100} {\pi N}$
0%
$\cfrac {6} {\pi N}$
0%
$\cfrac {25} {4N\pi}$

Explanation

Critical density of a gas

$= 0.44 gcm^{-3}$

Molar mass of

$CO_2$ = 44 g/mole

$Density = mass /volume$

Mass of N molecules = 44 g of

$CO_2$

$N =6.022\times 10^{23}$

Mass of 1 molecule

$= \dfrac{44}{N}$

The volume of 1 molecule

$=\cfrac {4}{3}\pi r^3$

Putting the values,

$0.44 = \dfrac{\dfrac{44}{N}}{\cfrac {4}{3}\pi r^3}$

$r^3 = \dfrac{100}{\pi N}$

The Statue of Liberty is made of

$2.0\times 10^5$ lbs of copper sheets bolted to a framework

$(1 lb =454 g)$ . How many atoms of copper are there on the statue?

(Atomic weight:

$Cu=63.5$ ).

0%
$2.1\times 10^{27}$
0%
$8.6\times 10^{29}$
0%
$4.3\times 10^{26}$
0%
$8.6\times 10^{26}$

Explanation

Given,

mass of copper

$= 2 \times 10^5 lbs$

$= 2 \times 10^5 \times 454 g$

Molecular wt. of copper

$= 63.5 g/moles$

$\therefore$ No. of moles of copper

$= \dfrac{2 \times 10^5 \times 454}{63.5}$

Thus, no. of atoms of copper on the statue

$= \dfrac{2 \times 10^5 \times 454}{63.5} \times 6.023 \times 10^{23}$

$= 8.6 \times 10^{29}$ atoms

Hence, the correct option is (B)

Number of

$HCl$ molecules present in

$10$ mL of

$0.1 N$

$HCl$ solution is

0%
$6.022 \times 10^{23}$
0%
$6.022 \times 10^{22}$
0%
$6.022 \times 10^{21}$
0%
$6.022 \times 10^{20}$

One gram molecule of any gas at

$NTP$ occupies

$22.4\ L$ . This fact was derived from:

0%
Dalton's theory
0%
Avogardro's hypothesis
0%
Berzelius hypothesis
0%
Law of gaseous volume

Explanation

According to Avogadro's law, 1 mole of every gas occupies 22.4L at

$NTP$ .

From ideal gas equation,

$PV=nRT$ . . . . . . .(1)

$NTP$ ,

$T=273K$

$P=1atm$

$n=1$

$R=0.0821atm L/K mol$

from equation (1),

$V=\dfrac{nRT}{P}$

$V=\dfrac{1\times 0.0821\times 273}{1}=22.4L$

The correct option is B.

The weight of

$H_{2}C_{2}O_{4},2H_{2}O$ required to prepare 500 ml of 0.2 N solution is:

0%
1.26 g
0%
6.3 g
0%
1.575 g
0%
3.15 g

A gaseous mixture contains

$70\%N_{2}$ and a

$30\%$ unknown gas, by volume. If the average molecular weight of gaseous mixture is

$37.60$ , the molecular weight of unknown gas is:

0%
$42.2$
0%
$60$
0%
$40$
0%
$50$

Explanation

Average

$M.M =\displaystyle \dfrac{\sum \%\ abundance \times M.M}{\sum \%\ abundance }$

$\therefore 37.60=\dfrac{0.70 \times 28+0.30 \times M_{unknown}}{0.70+0.30}$

$\dfrac{37.60-19.60}{0.30}= M_{unknown}$

$\therefore M_{unknown}=60\ gm$

Hence, option

$B$ is correct.

$12$ gm of an alkaline earth metal gives

$14.8$ g of its nitride. The atomic mass of metal is:

0%
$12$
0%
$24$
0%
$20$
0%
$40$

Explanation

Let the alkaline earth metal be

$A$ .

$\therefore 3A+N_{2}\rightarrow A_{3}N_{2}$

$\Rightarrow \dfrac{Molecular\ mass\ of\ compound}{Atomic\ mass\ of\ metal}$

$=\dfrac{Wt.\ of\ compound}{Wt.\ of\ metal}$

Let the atomic mass of metal be

$x$ .

$\rightarrow$ Molecular mass of compound

$=3x+28$

$\Rightarrow \dfrac{3x+28}{3x}=\dfrac{14.8}{12}$

$\Rightarrow x=40$

A gas mixture contains

$50\%$ helium and

$50\%$ methane by volume. What is the percent by weight of methane in the mixture?

0%
$19.97\%$
0%
$20.05\%$
0%
$50\%$
0%
$80.03\%$

Explanation

Solution contain

$He+CH_4$

$\%$ by volume is same as

$\%$ by moles.

Molecular wt. of

$He=4\ g/mol$

Molecular wt. of

$CH_4=16\ g/mol$

Assuming 1 mole of

$He$ and 1 mole

$CH_4$

$\%\ wt$ of

$CH_4=\dfrac{wt\ of\ CH_4}{Total\ wt}\times 100=\dfrac{16}{4+16}\times 100=80.0\%$

The solubility of potassium chloride in water at

$20^oC$ is

$34.7g$ in

$100g$ of water. The density of solution is

$1.3g/ml$ . Calculate the % of mass/volume concentration of potassium chloride.

0%
$25.76\%$
0%
$33.49\%$
0%
$24.7\%$
0%
$1.3\%$

Number of

$g$ of oxygen in

$32.2\ g\ Na_{2}SO_{4}\ 10H_{2}O$ is:[

$Mol. w t=322$ ]

0%
$16.0$
0%
$2.24$
0%
$18.0$
0%
$22.4$

Explanation

Number of gm of oxygen in 32.2 g

$Na_2SO_4.10H_2O$ is No. of oxygen atoms in one molecule =4+10=14

Thus moles of oxygen in n moles of compound=14n

Molar mass of

$Na_2SO_4.10H_2O$ =46+32+64+180=322g/mol

Thus

$32.2g= \dfrac{32.2}{322}=0.1\ mols$

Thus moles of

$O=0.1\times14=1.4$

So weight is

$16\times 1.4=22.4g$

Hence, the correct option is

$D$

$\overset { 14 }{ \underset { 7 }{ N } }$ if mass attributed to electron were doubled & the mass attributed to protons were halved, the atomic mass would become approximately:-

0%
Halved
0%
Doubled
0%
Reduced by 25%
0%
Remain same

The weight of 1 mole of a gas of density is

${\text{0}}{\text{.1784}}{\kern 1pt} {\kern 1pt} {\text{g}}{\kern 1pt} {{\text{L}}^{ - 1}}{\text{at}}{\kern 1pt} {\text{NTP}}{\kern 1pt} {\text{is}}{\kern 1pt} -$

0%
0.1784 g
0%
1 g
0%
4 g
0%
4 amu

Find the concentration of glucose in blood which have osmotic pressure

$\pi = 7.7 \ atm$ at

$T = 25^oC$

0%
$0.31M$
0%
$0.45M$
0%
$0.56M$
0%
$0.89M$

Explanation

$\pi = CRT$

$7.7 = C \times 0.082 \times 298$

$C = \dfrac{7.7}{24.44} \times 100 = 0.31$

The crystalline salt

$Na_{2}SO_{4}.xH_{2}O$ on heating loses 55.9% of its weight. The formula of the crystalline salt is :

0%
$Na_{2}SO_{4}. 5 H_{2}O$
0%
$Na_{2}SO_{4}. 7H_{2}O$
0%
$Na_{2}SO_{4}. 2H_{2}O$
0%
$Na_{2}SO_{4}. 10H_{2}O$

Calculate the concentration in

$m/v \%$ (in (mg/mL) of KCl solution which contains 0.0245 gm of KCl in 35 mL of solution.

0%
40
0%
60
0%
70
0%
30

The energy of an electron in

$n^{th}$ orbit of hydrogen atom is:

0%
$\dfrac { 13.6 }{ { n }^{ 4 } } eV$
0%
$\dfrac { 13.6 }{ { n }^{ 3 } } eV$
0%
$\dfrac { 13.6 }{ { n }^{ 2 } } eV$
0%
$\dfrac { 13.6 }{ { n } } eV$

$5\%$ solution of sucrose is isotonic with

$3\%$ solution of a solute find molecular weight of that solute?

0%
$54 g$
0%
$108 g$
0%
$216 g$
0%
$65 g$

Explanation

$5\%$ Glucose solution is isotonic with

$3\%$ of unknown solute

$\therefore$ conc. of glucose

$=$ conc. of solute

$\dfrac{5}{180} = \dfrac{3}{M}$

$M = \dfrac{3 \times 180 }{5}= 108 g$

Therefore, the option is B.

3 g of a solute was dissolved in 9 g of water. Find the percentage mass of solute

0%
25%
0%
33.33%
0%
18.7%
0%
20%

0.132 g of an organic compound gave 50 ml of

$N_2$ at NTP. The weight percentage of nitrogen in the compound is close to

0%
15
0%
20
0%
48.9
0%
47.34

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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