MCQ Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 6

Calculate the number of iron atoms in a piece of iron weighing

$2.8 g$ . (Atomic mass of iron

$=56$ )

0%
$30.11\times { 10 }^{ 23 }$ atoms
0%
$3.11\times { 10 }^{ 23 }$ atoms
0%
$3.0115\times { 10 }^{ 22 }$ atoms
0%
$301.1\times { 10 }^{ 23 }$ atoms

Explanation

Given that, weight of iron =2.8 g

Atomic mass= 56

moles of iron

$= \dfrac{\text{wt. of iron}}{\text{atomic mass}}=\dfrac{2.8}{56}=0.05\ moles$

1 mole

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

0.05 moles

$= 6.022 \times 0.05 \times 10^{23}\ atoms=3.0115 \times 10^{22}$ atoms

Option C is correct.

State whether true or false:

1 g of

$^{12}C$ contains

$6.022 \times 10^{23}$ atoms of the isotope.

0%
True
0%
False

Explanation

Because according to Avegadro's law,

$1$ mole of

$C$ contains

$6.022\times { 10 }^{ 23 }$ atoms and

$1$ mole of

$C$ weighs

$12gm$

$\therefore$

$12g$ of

$C$ weighs

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

8 g of a solute is dissolved in 92 g of solvent. If the density of solution is 2 g ml

$^{-1}$ , find the weight-volume percentage.

0%
10 g ml $^{-1}$
0%
12 g ml $^{-1}$
0%
16 g ml $^{-1}$
0%
15 g ml $^{-1}$

Explanation

given

$8\,g$ solute

$92\,g\,$ solvent

solution

$= 100\,g$

density

$= \dfrac{mass}{volume} = 2$

$\dfrac{100}{volume} = 2$

volume

$= 50\,ml$

weight

$-$ volume

$\% = \dfrac{weight\,of\,solute}{volume\,of\,sol^{n}}\times 100 = \dfrac{8}{50}\times 100$

$= 16\,g\,ml^{-1}$

1097196_516057_ans_c199aac24be04cd59fafd9d856f3cc76.jpg

Solubility of a solute is

$S$ . Find its percentage by weight.

0%
$\dfrac{200S}{S + 100}$
0%
$\dfrac{50S}{S + 100}$
0%
$\dfrac{100S}{S + 100}$
0%
None of these

Explanation

Solubility of solute is S means that

S grams per 100g solvent.

So weight of solution

$= (S+100)$ grams

weight %

$= \dfrac{weight\,of\,solute}{weight\,of\,solution}\times 100$

$= \dfrac{100\,S}{5+100}$

1100190_517027_ans_8f86df57a21649a2a1c8e5406c4e958a.jpg

A solution with a weight percentage of sodium hydroxide of 9% is prepared by adding the following mass of water (in grams) to 300 grams of that solution with a weight percent of

$NaOH$ of 15%.

0%
100 g
0%
150 g
0%
200 g
0%
250 g

Explanation

15 =

$\dfrac{NaOH weight}{(total weight)}\times100$

in this eqution total weight is 300gm so NaOH weight is 45 gm

in this question when x gm of water add in this solution we will get weight percent of NaOH is 9%

so the eqution is

9 =

$\dfrac{45}{300+x}\times100$

in sove this eqution x is 200gm

so the answer is 200gm

answer is C

How many mL of sulphuric acid of density 1.84 g mL

$^{-1}$ containing 95.6 mass % of

$H_2SO_4$ should be added to one litre of 40 mass % solution of

$H_2SO_4$ of density 1.31 g mL

$^{-1}$ in order to prepare 50 mass % solution of sulphuric acid of density 1.40 g mL

$^{-1}$ ?

0%
66.2 ml
0%
55.2 ml
0%
95.5 ml
0%
105.2 ml

Explanation

Let mass of solution be 100 gm
Therefore amount of solute in it = 95.6 gm
Volume of solution =

$\dfrac{mass}{density}$ =

$\dfrac{100}{1.84}$ = 54.3478 ml

Volume = 1 L =1000 ml
Density = 1.31 g/ml
Mass of solution = 1310 g
Mass % of solute = 40 %

Let mass of solute in it be X g

Mass % =

$\dfrac{mass of solute} {mass of solution} \times100$

40 =

$\dfrac{X}{1300}\times100$

X =

${40\times1310}$ = 524 gm

For solution to make 50 % mass percent and density 1.40 g/ml
Volume of solution =1400 ml
Amount of solute present should be,

Y gm =

$\dfrac{50\times1400}{100}$ = 700 g

Thus
Amount of solute needed to be added to the 40 % to make it 50% by mass =700-524 = 176 g

Volume of H2SO4 of density 1.84 g/ml required to be added to 40 % =

$\dfrac{mass}{density}$ =

$\dfrac{176}{1.84}$ =95.6521 ml

so the answer is C

What weight of sodium contains the same number of atoms as those in

$8$ grams of oxygen?

0%
$10.5 g$
0%
$13.5 g$
0%
$11.5 g$
0%
$14.2 g$

Explanation

Atomic weights of sodium and oxygen are 23 g/mol and 16 g/mol respectively.

8 g oxygen

$\displaystyle = \dfrac {8 \ g}{16 \ g/mol}= 0.5 \$ moles of oxygen atoms.

Number of moles of sodium

$\displaystyle =$ number of moles of oxygen atoms

$\displaystyle =$ 0.5.

Weight of sodium

$\displaystyle = 0.5 \ mol \times 23 \ g/mol = 11.5 \ g$ .

Note: Since sodium and oxygen samples have same number of atoms, they will have same number of moles.

A student has only 200 g of valuable solvent and wishes to make a solution of 20% by weight from solute

$A$ . How many grams of

$A$ is weighed out?

0%
50 grams
0%
250 grams
0%
20 grams
0%
30 grams

Explanation

$200 g$ solvent

$Solute = 20\%$

$\dfrac{weight\, of\, solute}{total\, weight}=\dfrac{20}{100}$

$\dfrac{x}{200+x}=\dfrac{1}{5}$

$x=50$ grams.

The value of the Avogadro constant is:

0%
$6.022 \times 10^{13}$
0%
$6.022 \times 10^{22}$
0%
$6.022 \times 10^{23}$
0%
$6.022 \times 10^{24}$

Explanation

Correct Answer: Option C

Explanation:

The value of the Avogadro constant is $6.022 \times 10^{23}$ .

It is a number of atoms or molecules or ions or particles present in one mole of a substance.

Hence, the correct answer is option $C$ .

If 11g of oxalic acid are dissolved in 500 mL of solution (density= 1.1 g

$mL^{-1}$ ), what is the mass % of oxalic acid in solution?

0%
1%
0%
2%
0%
3%
0%
4%

Explanation

Mass of Oxalic acid (solute) = 11g

Mass of Solution = 500ml

Density = 1.1 g/ml

We know that

Mass % of solute =

$\dfrac{Mass of solute × 100}{Massofsolution}×50$

Mass of solution =

$\text {Volume of solution} \times \text{Density of solution}$

Mass of solution =

$500\times 1.1=50\times 11g$

Mass % of the solute =

$\dfrac{11\times 100} {11\times 50}=2%$

the answer is B

Calculate the percentage composition of nitrogen in urea (

$H_2NCONH_2$ ).

0%
46.6% $N$
0%
43.6% $N$
0%
44.6% $N$
0%
42.6% $N$

Explanation

Molecular weight of

${ H }_{ 2 }{ NCONH }_{ 2 }=4+28+12+16=60g$

$60g$ of urea contains

$28g$ of

$N$

$\therefore$ % of

$N=\dfrac { 28 }{ 60 } \times 100=46.6$ %

The number of moles of NaCl in 3 litres of 3 M solution is:

0%
1
0%
3
0%
9
0%
27

Explanation

molarity is equal to the number of moles of solute upon the volume of solution in liter.

$MOLARITY = \dfrac{\text{number of moles }}{\text {volume of solution}}$

number of moles =

$3\times 3$ = 9

so the answer is 9 C option

What is the percentage composition of the elements in ammonia (

$NH_3$ )?

$(N = 14,\ H = 1)$

0%
$N= 82.35$ %, $H=17.65$ %
0%
$N= 81.35$ %, $H=18.65$ %
0%
$N= 80.35$ %, $H=19.65$ %
0%
$N= 42.35$ %, $H=57.65$ %

Explanation

Molecular weight of

${ NH }_{ 3 }=17g$

$17g$ of

${ NH }_{ 3 }$ contains

$14g$ of

$N$ and

$3g$ of

$H$

$\therefore$ % of

$H=\dfrac { 3 }{ 17 } \times 100=17.65$ %

$\therefore$ % of

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

What is the mass of a mole of water containing 50% of heavy water

$(D_2O)$ ?

0%
$19\ g$
0%
$18\ g$
0%
$20\ g$
0%
$21\ g$

Explanation

A mole of water containing 100 % of

$\displaystyle H_2O$ weighs

$\displaystyle 2(1)+16=18$ g.

A mole of water containing 100 % of

$\displaystyle D_2O$ weighs

$\displaystyle 2(2)+16=20$ g.

A mole of water containing 50 % of

$\displaystyle H_2O$ and 50 % of

$\displaystyle D_2O$ weighs

$\displaystyle \dfrac { 18+20}{2}=19$ g.

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 =

$\dfrac{\text{mass of glucose}}{\text{mass of glucose +mass of water}} \times 100$

$\dfrac{5.64}{5.64+60} \times 100$

= 8.59 percent

answer is A

What are the percentage compositions of hydrogen and oxygen in water

$(H_2O)$ ?

$(H =1,\ O = 16)$

0%
11.1, 88.9
0%
11.2, 88.8
0%
11, 89
0%
11.3, 88.7

Explanation

Molecular weight of

${ H }_{ 2 }O=2+16=18g$

$18g$ of

${ H }_{ 2 }O$ contains

$2g$ of

$H$ and

$16g$ of

$O$

$\therefore$ % of

$H=\dfrac { 2 }{ 18 } \times 100=11.1$ %

% of

$O=\dfrac { 16 }{ 18 } \times 100=88.9$ %

$250$ g sample of a hydrated salt was heated at

${110}^{o}C$ until all water was driven off. The remaining solid weighed

$160$ g. From these data, the percent of water by weight in the original sample can be correctly calculated as :

0%
$\cfrac { 160 }{ 340 } \times 100$ %
0%
$\cfrac { 90 }{ 340 } \times 100$ %
0%
$\cfrac { 160 }{ 250 } \times 100$ %
0%
$\cfrac { 90 }{ 250 } \times 100$ %
0%
$\cfrac { 90 }{ 160 } \times 100$ %

Explanation

The percent of water by weight in the original sample can be correctly calculated as :

$= \cfrac { \text { mass of water } }{ \text { mass of hydrated sample } } \times 100$

$= \cfrac { \text { mass of hydrated sample } - \text { mass of anhydrous sample } }{ \text { mass of hydrated sample } } \times 100$

$= \cfrac { 250 -160 }{ 250 } \times 100$

$= \cfrac { 90 }{ 250 } \times 100$

14 g of nitrogen contains

$3.01 \times 10^{23}$ nitrogen molecules.

0%
True
0%
False

Explanation

1 mole nitrogen contains

$6.02\times10^{23}$ nitrogen molecules.

Thus 14 gram nitrogen will contain

$14/28.Na=0.5Na=3.01\times10^{23}$ nitrogen molecules. ( Molecular weight of nitrogen is 28)

If 'M' is the molecular weight of a gas, what volume in

$l$ at STP would be occupied by M/4 g of that gas?

0%
11.2
0%
2.24
0%
5.6
0%
22.4

Explanation

If 'M' is the molecular weight of a gas, then

$\displaystyle M/4$ g corresponds to

$\displaystyle 1/4$ moles.

At STP,

$1$ mole of gas occupies a volume of

$22.4$ L.

At STP,

$\dfrac 14$ mole of gas occupies a volume of

$\displaystyle \dfrac {22.4}{4}=5.6$ L.

Hence, option C is correct.

What is the percent composition by mass of aluminium in a compound of aluminium and oxygen if the mole ratio of

$Al : O$ is

$2:3$ ?

0%
$37$ % $Al$
0%
$47$ % $Al$
0%
$53$ % $Al$
0%
$63$ % $Al$
0%
$74$ % $Al$

Explanation

In a compound of aluminium and oxygen the mole ratio of

$Al : O$ is

$2:3$
The atomic masses of

$Al$ and

$O$ are 27.0 g/mol and 16 g/mol respectively.
2 moles of Al

$= 2 \times 27.0 = 54.0$ g
3 moles of O

$=3 \times 48.0$
Hence, the percent composition by mass of aluminium is

$\dfrac {54.0}{54.0+48.0} \times 100= 53$ %.

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. How many number molecules does container B have?

0%
2N
0%
4N
0%
N
0%
8N

What is the molar mass of ammonium carbonate

${({NH}_{4})}_{2}{CO}_{3}$ ?

0%
$48\ g/mol$
0%
$96\ g/mol$
0%
$82\ g/mol$
0%
$78\ g/mol$
0%
$192\ g/mol$

Explanation

$(NH_4)_2CO_3$ is the chemical formula of ammonium carbonate.

$N = 14 \times (2) = 28$

$H = 1 \times (4 \times 2) = 8$

$C = 12 \times 1 = 12$

$O = 16 \times 3 = 48$

Molar mass

$= 28 + 8 + 12 + 48 = 96\ g/mol$

64 g of sulphur dioxide occupies

$22.4\ L$ volume at STP.

0%
True
0%
False

Explanation

Molecular mass of

$SO_2$ is 64.

So, 1 mole (64 grams) of sulphur dioxide will occupy

$22.4\ L$ at STP. ( Avogadro's law)

Hydrogen, oxygen and carbon dioxide are taken in containers of

$2 l$ volume each. Compare the ratio of the number of molecules of the three gases respectively, under same conditions of temperature and pressure.

0%
1:8:22
0%
1:1:1
0%
1:16:44
0%
1:8:44

Explanation

Using ideal gas law: PV =nRT

n =

$\dfrac{V}{RT}$

here P, R and T= Constant, V = 2l

hence number of moles also equal

the ration of moles equal to 1:1:1

answer is B

$2K(s)+2{H}_{2}O(l)\rightarrow 2KOH(aq)+{H}_{2}(g)$
If

$3.0$ moles of potassium react with excess water, what volume of hydrogen gas will be produced?

0%
$1.5L$
0%
$22.4L$
0%
$67.2L$
0%
$33.6L$

Explanation

Given,

$2K(s) + 2H_2O(l) \rightarrow 2KOH(aq) + H_2 (g)$

$2\space moles$ of Potassium on reaction gives

$1 \space mole$ of Hydrogen [ 22.4 at STP]

$2\space moles$ of

$K \rightarrow 22.4 \space Litres$ of

$H_2$

$2\space moles$ of

$K = \dfrac{3\times 22.4}{2} = 33.6 \space Litres$ of

$H_2$

So,

$33.6\space Litres$ of Hydrogen are produced.

What is the atomic mass of sodium?

0%
$22\ g/mol$
0%
$23\ g/mol$
0%
$11\ g/mol$
0%
$20\ g/mol$

What is the relative atomic mass of cadmium?

0%
112.411 amu
0%
20.00 amu
0%
21.00 amu
0%
20.43 amu

The father of modern chemistry is:

0%
Priestley
0%
Lavoisier
0%
Dalton
0%
Mendeleev

Explanation

Antoine Lavoisier is known as the father of modern chemistry. Lavoisier wrote the book "Elements of Chemistry" (1787). He compiled the first complete (at that time) list of elements, discovered and named oxygen and hydrogen, helped develop the metric system, helped revise and standardize chemical nomenclature, and discovered that matter retains its mass even when it changes forms.

Hence, the correct option is

$B$

$5.0g$ sample of copper(II) oxide containing an inert contaminant is analyzed and found to contain

$3.0g\ Cu$ . Using molar masses of

$64g/mol$ for copper, and

$16g/mol$ for oxygen, determine the percent purity of the copper(II) oxide sample.

0%
$80$ %
0%
$75$ %
0%
$60$ %
0%
$50$ %

Explanation

Given, Mass of

$CuO$ sample

$= 5\space g$

Mass of

$Cu$ found

$= 3\space g = \dfrac{3}{64} \space moles$ of Copper

So,

$\dfrac{3}{64} \space moles$ of

$CuO$ must be present

Mass of

$\dfrac{3}{64} \space moles$ of

$CuO = \dfrac{3}{64} [64 + 16] = 3.75 \space g$

So, percentage of Purity

$= \dfrac{3.75}{5} \times 100 = 75\%$

Use the figure below to answer the question that follows.
Which of the four molecules has the highest percentage of nitrogen, by mass?

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

Explanation

If the other atoms of molecules have comparitively less atomic mass then percentage of central atom is more.

$II \Rightarrow N_2H_4$ has Hydrogen atoms which has very less atomic mass (1) compared to Nitrogen (14). So,

$N_2H_4$ has highest percentage of Nitrogen.

A student would like to use gravimetric analysis to determine the percent by mass of magnesium chloride in a contaminated mixture containing magnesium chloride.
Which of the following compounds would be the BEST substance to react the contaminated mixture with to determine the mass percent?

0%
silver nitrate.
0%
Potassium chloride.
0%
Sodium acetate.
0%
Lithium sulfate.

How many grams are there in a

$7.0$ mole sample of sodium hydroxide?

0%
$40.0g$
0%
$140.0\ g$
0%
$280.0\ g$
0%
$340.0\ g$
0%
$420.0\ g$

Explanation

The formula of sodium hydroxide is

$NaOH$

Atomic weight of

$Na$ =

$23\ {g}/{mol}$

Atomic weight of

$O$ =

$16\ {g}/{mol}$

Atomic weight of

$H$ =

$1\ {g}/{mol}$

$\therefore$ molecular weight of

$NaOH$ = atomic mass of

$Na$ + atomic mass of

$O$ + atomic mass of

$H$

$23 + 16 + 1 = 40\ {g}/{mol}$

Therefore, 1 mole of

$NaOH = 40\ g/mol$

7 moles of

$NaOH = 7 \ mole \times 40\ g/mol= 280g$

Hence, option

$B$ is correct.

A hydrate sample of magnesium chloride is analyzed and found to contain

$53.18$ % water.
How many water molecules are found in the formula for this hydrate?

0%
Two
0%
Four
0%
Six
0%
Eight

Explanation

$\begin{array}{l}\text { Let, the hydrated sample contains } x \text { molecules of } \mathrm{H}_{2} \mathrm{O} \text { . } \\\text { so, given sample becomes } \mathrm{MgCl}_{2} \cdot x\mathrm{H}_{2}\mathrm{O} \text { . } \\\text { Now, it is given that } 53.18 \% \text { of the formula is } \mathrm{H}_{2} \mathrm{O} \text { . }\end{array}$

$\begin{array}{l}\text { Thus, } \dfrac{x \times 18}{(18x+95)}=\dfrac{53.18}{100 .} \Rightarrow x=6 \\\text { Hence, there are six water molecules are found in this hydrate. }\end{array}$

What would be the approximate weight of

$1.204\times 10^{24}$ bromine atoms?

0%
80 grams
0%
120 grams
0%
160 grams
0%
180 grams
0%
200 grams

Explanation

The atomic weight of bromine is 80.

So, 80-gram bromine contains Na atoms.

So,

$1.204\times10^{24}$ bromine atoms will weigh:

$80\div 6.02\times10^{23}\times 1.204\times10^{24}=160\ gram$

Which pair of substances share the same mass percent of each element?

0%
$NaOCl$ and $OCl^{-}$
0%
$CO$ and $CO_{2}$
0%
$C_{6}H_{12}O_{6}$ and $C_{3}H_{6}O_{3}$
0%
None of these pairs of substances share the same mass percent.

Explanation

When the ratio of number of atoms in a molecule are same then the percentage of each element will be same.

Here, In

$C_6H_{12}O_6$

$C:H:O = 6:12:6 = 1:2:1$

$C_3H_{6}O_3$

$C:H:O = 3:6:3 = 1:2:1$

So,

$C_6H_{12}O_6$ and

$C_3H_{6}O_3$ has same percent of element.

A mixture contains

$5.0\ g$ of compound

$X$ and

$20.0\ g$ of compound

$Y$ .What is the percent by mass of compound

$X$ ?

0%
$75$
0%
$25$
0%
$80$
0%
$20$

Explanation

Given,

Mass of X

$= 5g = m_x$

Mass of Y

$= 20g = m_y$

$\%$ Mass of X is defined as ratio of mass of X and total mass of mixture.

$\%$ Mass of X

$= (\dfrac{m_x}{m_x + m_y}) \times 100 = (\dfrac{5}{5 + 20}) \times 100 = 20\%$

How many atoms of hydrogen are present in

$7.8\ g$ of

$Al{(OH)}_{3}$ ?

0%
$6.0\times {10}^{22}$
0%
$1.8\times {10}^{23}$
0%
$1.7\times {10}^{23}$
0%
$5.1\times {10}^{22}$

Explanation

Molar mass of

$Al(OH)_3 = 27 + 3\times 16 + 3\times 1 = 78\space g$

So, no. of moles of

$Al(OH)_3 = \dfrac{7.8}{78} = 0.1\space moles$

$1\space mole$ of

$Al(OH)_3$ has

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

So,

$0.1\space moles$ has

$6.023 \times 10^{22}$ atoms.

What could be better than a dozen

$(12)$ donuts? How about a baker's dozen

$(13)$ of donuts? Another large unit of measurement is known as Avogadro's number

$(6.022\times 10^{23})$ .
What is TRUE about Avogadro's number?

0%
Avogadro's number is the number of particles in one mole of any element.
0%
The molar mass of a substance will contain $6.022\times 10^{23}$ molecules.
0%
There are $6.022\times 10^{23}g$ of carbon in $12$ mol of $C - 12$ .
0%
$6.022\times 10^{23}$ atoms of any element will have the exact same mass regardless of the identity of the element.

Explanation

Avagadro constant i.e.

$6.023 \times 10^{23}$ particles is the number of constituent particles, usually atoms or molecules, that are contained in the amount of substance given by

$1\space mole$ .

What is the correct unit for the following solution?

$10.5\ g$ of

${C}_{7}{H}_{8}O$ in

$100\ g$ of water as a topical treatment for poison ivy.

0%
$w/w$
0%
$m$
0%
$M$
0%
$v/v$

What is the mass of

$6.022\times { 10 }^{ 23 }$ formula units of

${({NH}_{4})}_{2}{SO}_{4}$ ?

0%
$234.11\ g$
0%
$132.11\ g$
0%
$210.29\ g$
0%
$342.14\ g$

Explanation

$1\space mole$ of $(NH_4)_2SO_4$ has $6.023 \times 10^{23}$ formula units.

So, the mass of $6.023 \times 10^{23}$ formula units of $(NH_4)_2SO_4$ . is molar mass of $(NH_4)_2SO_4$

Molar mass of $(NH_4)_2SO_4$ $= 2(14 + 4) + 32 + 4\times 16 = 132 \space g$

A compound contains 38.8% C, 16.0% H and 45.2% N. The formula of the compound would be

0%
$\mathrm { CH } _ { 3 } \mathrm { NH } _ { 2 }$
0%
$\mathrm { CH } _ { 3 } \mathrm { CN }$
0%
$\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { CN }$
0%
$\mathrm { CH } _ { 2 } ( \mathrm { NH } ) _ { 2 }$

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

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

${X}_{(g)}$ by energy

$\Delta H$ . The IE of

$X$ is

0%
$\cfrac{2\Delta H}{{N}_{A}}$
0%
$\cfrac{2{N}_{A}}{\Delta H}$
0%
$\cfrac{\Delta H}{2{N}_{A}}$
0%
$\cfrac{{N}_{A}}{\Delta H}$

The image below represents the formation of potassium bromide,

$KBr$ . Which law is represented by this reaction?

0%
Law of Conservation of Mass
0%
Law of Multiple Proportions
0%
Law of Definite Proportions
0%
Law of Stoichiometry

Using the mass spectrum provided below:
Determine the relative atomic mass and identify the element.

0%
$Be, 9.0122\ amu$
0%
$B, 10.199\ amu$
0%
$C, 12.011\ amu$
0%
$N, 14.007$

Explanation

The relative atomic mass

$\displaystyle =\dfrac { [ \text { atomic mass of }^{10}X \times \text {abundance of }^{10}X ]+ [ \text { atomic mass of }^{11}X \times \text {abundance of }^{11}X ] }{ \text {abundance of }^{10}X + \text {abundance of }^{11}X } \\ \:=\dfrac { [ \text {10 } \times \text {80.1 } ]+[ \text { 11} \times \text {19.9 } ] }{\text {80.1 } + \text { 19.9} } = \text { 10.199 amu}$

10.199 is close to atomic mass of B which is 10.8. Hence, our element is boron.

The atomic mass of copper is

$64\ amu$ , while the atomic mass of oxygen is

$16\ amu$ .
Let

$= copper$ and

$= oxygen$ .
Which of the following would represent a compound that was

$50$ % copper, by mass?

Explanation

Given, compound has

$50\%$ copper, by mass.

Let Molar mass of compound

$= M$

$\Rightarrow$ Mass of copper

$\dfrac{50}{100} \times M = 0.5 M$

No. of moles of copper

$= \dfrac{0.5M}{64} ;$

No. of moles of oxygen

$= \dfrac{0.5M}{16} ;$

Ratio of copper to oxygen

$1:4$

$\Rightarrow CuO_4$

If a copper-containing compound contains

$4.999$ %

$Cu$ by mass and there are

$2$ copper atoms per formula unit, what is the molar mass of the compound?

0%
$2547\ g/mol$
0%
$47.50\ g/mol$
0%
$31.75\ g/mol$
0%
$1274\ g/mol$

Explanation

Let molar mass of compound

$= M$

$1\space mole$ of compound there are two Copper atoms i.e.

$128\space gms$ of Copper.

Given,

$4.999\%$ ofCopper by mass

So,

$(\% \text{ of Copper}) \times \text{Molar mass} = \text{Mass of Copper}$

$\Rightarrow (\dfrac{4.999}{100}) \times M = 128 \space gms$

$\Rightarrow M = \dfrac{12800}{4.99} = 2547 \space g/mol$

How many molecules of methane are present in a sample that contains

$1.0\times { 10 }^{ 10 }$ moles of methane?

0%
$6.0\times { 10 }^{ 23 }$
0%
$1.0\times { 10 }^{ 33 }$
0%
$6.0\times { 10 }^{ 33 }$
0%
$6.0\times { 10 }^{ 24 }$

Explanation

$1\space mole$ of metane has $6.023 \times 10^{23}$ molecules of methane.

$6.023 \times 10^{23}$ molecules

$10^{10}\space moles \rightarrow x$

$(6.023 \times 10^{23})\times 10^{10} =$ $6.023 \times 10^{33}$ molecules

$10^{10}\space moles$ has approximately $6.0 \times 10^{33}$ molecules

How many grams of

$CaWO_4$ would contain the same mass of tungsten that is present in 569g of $$FeWO_4

? (W=184)

0%
$298$ g
0%
$540$ g
0%
$487$ g
0%
$474$ g

Explanation

Let the mass of

$CaWO_4$ be w g. As given,

Mass of W in w g of

$CaWO_4$ = mass of W in 569g of

$FeWO_4$

Moles of W in

$CaWO_4 \times$ at. mass of W = Moles of W in

$FeWO_4 \times$ at. mass of W

As both

$CaWO_4$ and

$FeWO_4$ contains 1 atom of W each,

∴ moles of

$CaWO_4 \times$ at. mass of W = Moles of

$FeWO_4 \times$ at. mass of W

$\dfrac{w}{288} \times 184$ =

$(569/304) × 184$

$W = 539.05 g.$

What is the percent

$X$ by mass for compound

$XY_{2}$ if compound

$XY$ is

$74$ %

$X$ by mass?

0%
$25$ %
0%
$40$ %
0%
$60$ %
0%
$75$ %

Explanation

Given

$\%$ of X in

$XY = 74\%$

So,

$\%$ Y in

$XY = 26\%$

$\Rightarrow \dfrac{\text{Mass of X}}{\text{Mass of X + Mass of Y}} = \dfrac{74}{100}$

$\Rightarrow 100 \text{ mass of X} = 74(\text{mass of X}) + 74(\text{mass of Y})$

$\Rightarrow 26 \text{ mass of X} = 74(\text{mass of Y})$

$\Rightarrow 13 \text{ mass of X} = 37(\text{mass of Y})$

$\%$ of X in

$XY_2 = \dfrac{\text{Mass of X}}{\text{(Mass of X) + 2(Mass of Y)}} \times 100 = \dfrac{\dfrac{37}{13}(\text{Mass of Y})}{\dfrac{63}{13}(\text{Mass of Y} )} \times 100 = \dfrac{37}{63} \times 100 = 60\%$

Equal masses of oxygen, hydrogen, and methane are taken in a container in identical condition. Find the ratio of volumes of the gases.

0%
1: 32 :2
0%
1 : 16: 2
0%
1 : 1: 1
0%
1: 16 :1

Explanation

The molecular weights of a given mixture of gases are

1. Oxygen = 32

2. Hydrogen = 2

3. Methane = 16

Since masses are equal, assume mass as x,

So By using formula, no of moles =

$\dfrac{mass }{molecular weight}$

$n_{O_2} = \dfrac{x}{32}$ :

$n_{H_2} = \dfrac{x}{2}$ :

$n_{CH_4} = \dfrac{x}{16}$

ratio is

$1 : 16 : 2$

CBSE Questions for Class 11 Medical Chemistry Some Basic Concepts Of Chemistry Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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