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Chemistry 1 Marks Question

Some Basic Concepts of Chemistry · Jharkhand Board (JAC) STD 11 Science (Jharkhand - English Medium). 48 questions.

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Question 11 Mark
Which one of the following will have the largest number of atoms?
1g Li(s)
Answer
$1 \mathrm{~g} \text { of } \mathrm{Li}(\mathrm{s})=\frac{1}{23} \mathrm{~mol} \text { of } \mathrm{Li}(\mathrm{s})$
$=\frac{6.022 \times 10^{23}}{197} \text { atoms of } \mathrm{Li}(\mathrm{s})$
$=0.86 \times 10^{23} \text { atoms of } \mathrm{Li}(\mathrm{s})$
$=86.0 \times 10^{21} \text { atoms of } \mathrm{Li}(\mathrm{s})$
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Question 21 Mark
What is the SI unit of mass? How is it defined?
Answer
The SI unit of mass is kilogram (kg). 1 Kilogram is defined as the mass equal to the mass of the international prototype of kilogram.
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Question 31 Mark
Convert the following into basic units:
15.15pm
Answer
$1 \mathrm{pm}=10^{-12} \mathrm{~m}$
$\therefore 15.1 \mathrm{pm}=15.15 \times 10^{-12} \mathrm{~m}=1.515 \times 10^{-11} \mathrm{~m}$
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Question 41 Mark
Convert the following into basic units:
28.7pm
Answer
$1 \mathrm{pm}=10^{-12} \mathrm{~m}$
$\therefore 28.7 \mathrm{pm}=28.7 \times 10^{-12} \mathrm{~m}=2.87 \times 10^{-11} \mathrm{~m}$
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Question 51 Mark
How many significant figures are present in the following?
5005
Answer
5005
There are 4 significant figures.
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Question 61 Mark
Convert the following into basic units:
25365mg
Answer
$1 \mathrm{mg}=10^{-3} \mathrm{~g}$
$25365 \mathrm{mg}=2.5365 \times 10^4 \times 10^{-3} \mathrm{~g}$
$\text { Now, }$
$1 \mathrm{~g}=10^{-3} \mathrm{~kg}$
$2.5365 \times 10 \mathrm{~g}=2.5365 \times 10 \times 10^{-3} \mathrm{~kg}$
$\therefore 25365 \mathrm{mg}=2.5365 \times 10^{-2} \mathrm{~kg}$
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Question 71 Mark
Calculate the amount of carbon dioxide that could be produced when
1 mole of carbon is burnt in 16g of dioxygen.
Answer
The balanced reaction of combustion of carbon can be written as:
$\ \ \ \ \text{C}_{\text{s}} \ \ \ \ + \ \ \ \ \text{O}_{2(\text{g})} \ \ \rightarrow \ \ \text{C}_{2(\text{g})} \\ \ \text{1 mole} \ \ \ \ \ \ \ \text{1 mole} \ \ \ \ \ \ \ \text{1 mole} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (32\text{g})\ \ \ \ \ \ \ \ (44\text{g})$
According to the question, only 16g of dioxygen is available.
Hence, it will react with 0.5 mole of carbon to give 22g of carbon dioxide.
Hence, it is a limiting reactant.
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Question 81 Mark
Which one of the following will have the largest number of atoms?
1g Na(s)
Answer
$1 \mathrm{~g} \text { of } \mathrm{Na}(\mathrm{s})=\frac{1}{23} \mathrm{~mol} \text { of } \mathrm{Na}(\mathrm{s})$
$=\frac{6.022 \times 10^{23}}{197} \text { atoms of } \mathrm{Na}(\mathrm{s})$
$=0.262 \times 10^{23} \text { atoms of } \mathrm{Na}(\mathrm{s})$
$=26.2 \times 10^{21} \text { atoms of } \mathrm{Na}(\mathrm{s})$
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Question 91 Mark
Calculate the amount of carbon dioxide that could be produced when
2 moles of carbon are burnt in 16g of dioxygen.
Answer
The balanced reaction of combustion of carbon can be written as:
$\ \ \ \ \text{C}_{\text{s}} \ \ \ \ + \ \ \ \ \text{O}_{2(\text{g})} \ \ \rightarrow \ \ \text{C}_{2(\text{g})} \\ \ \text{1 mole} \ \ \ \ \ \ \ \text{1 mole} \ \ \ \ \ \ \ \text{1 mole} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (32\text{g})\ \ \ \ \ \ \ \ (44\text{g})$
According to the question, only 16g of dioxygen is available.
It is a limiting reactant.
Thus, 16g of dioxygen can combine with only 0.5 mole of carbon to give 22g of carbon dioxide.
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Question 101 Mark
How many significant figures are present in the following?
208
Answer
208
There are 3 significant figures.
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Question 111 Mark
Which one of the following will have the largest number of atoms?
1g Au(s)
Answer
$1 \mathrm{~g} \text { of } \mathrm{Au}(\mathrm{s})=\frac{1}{197} \mathrm{~mol} \text { of } \mathrm{Au}(\mathrm{s})$
$ =\frac{6.022 \times 10^{23}}{197} \text { atoms of } \mathrm{Au}(\mathrm{s})$
$ =3.06 \times 10^{21} \text { atoms of } \mathrm{Au}(\mathrm{s})$
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Question 121 Mark
Which one of the following will have the largest number of atoms?
1g of $\mathrm{Cl}_2(\mathrm{g})$
Answer
1 g of $\mathrm{Cl}_2(\mathrm{g})=\frac{1}{71} \mathrm{~mol}$ of $\mathrm{Cl}_2(\mathrm{~g})$
(Molar mass of $\mathrm{Cl}_2$ molecule $=35.5 \times 2=71 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$=\frac{6.022 \times 10^{23}}{71}$ molecules of $\mathrm{Cl}_2(\mathrm{~g})$
$=0.0848 \times 10^{23}$ molecules of $\mathrm{Cl}_2(\mathrm{~g})$
$=8.48 \times 10^{21}$ molecules of $\mathrm{Cl}_2(\mathrm{~g})$
As one molecule of $\mathrm{Cl}_2$ contains two atoms of Cl .
Number of atoms of $\mathrm{Cl}=2 \times 8.48 \times 10^{21}=16.96 \times 10^{21}$ atoms of Cl .
Hence, 1 g of $\mathrm{Li}(\mathrm{s})$ will have the largest number of atoms.
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Question 131 Mark
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
$\mathrm{N}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NH}_{3(\mathrm{~g})}$
Will any of the two reactants remain unreacted?
Answer
$\mathrm{N}_2$ is the limiting reagent and $\mathrm{H}_2$ is the excess reagent.
Hence, $\mathrm{H}_2$ will remain unreacted.
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Question 141 Mark
The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
 
Mass of dinitrogen
Mass of dioxygen
(i)
14g
16g
(ii)
14g
32g
(iii)
28g
32g
(iv)
28g
80g
Which law of chemical combination is obeyed by the above experimental data?
Give its statement.
Answer
If we fix the mass of dinitrogen at 28g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32g, 64g, 32g, and 80g.
The masses of dioxygen bear a whole number ratio of 1 : 2 : 2 : 5.
Hence, the given experimental data obeys the law of multiple proportions.
The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.
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Question 151 Mark
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
$\mathrm{N}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NH}_{3(\mathrm{~g})}$
If yes, which one and what would be its mass?
Answer
Mass of dihydrogen left unreacted $=1.00 \times 10^3 \mathrm{~g}-428.6 \mathrm{~g}=571.4 \mathrm{~g}$.
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Question 161 Mark
How many significant figures are present in the following?
500.0
Answer
500.0
There are 4 significant figures.
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Question 171 Mark
How many significant figures are present in the following?
2.0034
Answer
2.0034
There are 5 significant figures.
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Question 181 Mark
How many significant figures are present in the following?
126,000
Answer
126,000
There are 3 significant figures.
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Question 191 Mark
How many significant figures are present in the following?
0.0025
Answer
0.0025
There are 2 significant figures.
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Question 201 Mark
Calculate the amount of carbon dioxide that could be produced when
1 mole of carbon is burnt in air.
Answer
The balanced reaction of combustion of carbon can be written as:
$\ \ \ \ \text{C}_{\text{s}} \ \ \ \ + \ \ \ \ \text{O}_{2(\text{g})} \ \ \rightarrow \ \ \text{C}_{2(\text{g})} \\ \ \text{1 mole} \ \ \ \ \ \ \ \text{1 mole} \ \ \ \ \ \ \ \text{1 mole} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (32\text{g})\ \ \ \ \ \ \ \ (44\text{g})$
As per the balanced equation, 1 mole of carbon burns in1 mole of dioxygen (air) to produce1 mole of carbon dioxide.
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Question 211 Mark
Boron occurs in nature in the form of two isotopes, $_5^{11}\text{B}\text{ and }_5^{10}\text{B},$ in ratio of 81% and 19% respectively. Calculate its average atomic mass.
Answer
Average atomic mass $=\frac{11\times81+10\times19}{100}=10.81$
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Question 231 Mark
What is volume of 17 g of $\mathrm{NH}_3$ at STP $(273 \mathrm{~K}, 1 \mathrm{~atm})$ ?
Answer
$22.4 \mathrm{~L}\left[\because 17 \mathrm{~g}\right.$ of $\left.\mathrm{NH}_3=1 \mathrm{~mole}\right][\mathrm{N}=14\mathrm{u} \mathrm{~H}=1 \mathrm{u}]$
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Question 241 Mark
What is the mass of 1 L of mercury in grams and in kilograms, if the density of liquid mercury is 13.6 g $\mathrm{cm}^{-3}$ ?
Answer
Mass $=$ Volume $\times$ Density $=1000 \mathrm{~cm}^3 \times 13.6 \mathrm{~g} \mathrm{~cm}^{-3}=13600 \mathrm{~g}=13.6 \mathrm{~kg}$.
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Question 261 Mark
What is relationship between properties of compound and its constituting elements.
Answer
A compound has properties different from it constituting atom.
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Question 271 Mark
Give an example of molecule in which the ratio of the molecular formula is six times the empirical formula.
Answer
The compound is glucose. Its molecular formula is $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$ while empirical formula is $\mathrm{CH}_2 \mathrm{O}$.
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Question 281 Mark
Give one example each of homogeneous and heterogeneous mixture.
Answer
Salt solution is an example of homogeneous mixture where as sugar and salt is heterogeneous mixture.
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Question 301 Mark
Calculate number of atoms in 32u of He.
Answer
$\text{Number of atoms}=\frac{\text{Given mass in u}}{\text{Molecular mass in u}}$
$=\frac{32\text{u}}{4\text{u}}=8\text{ atoms}.$
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Question 311 Mark
Classify the following as pure substances and mixtures: air, glucose, gold, sodium and milk.
Answer
  • Pure substances: glucose, gold, sodium.
  • Mixture: milk, air
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Question 321 Mark
A substance has molecular formula $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$. What is its empirical formula?
Answer
$\mathrm{CH}_2 \mathrm{O}$ is empirical formula.
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Question 331 Mark
Define accuracy.
Answer
It refers to the agreement of a particular value to the true value of the result.
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Question 391 Mark
Which bas more number of atoms?
1.0g of Na or 1.0g of Mg. [At mass of Na = 23u, Mg = 24u]
Answer
1.0g of Na.
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Question 421 Mark
What is AZT? Mention its use in medical science.
Answer
AZT is Azidothymidine. It is used to treat AIDS.
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Question 431 Mark
Suppose the length of a cardboard has been reported to be 31.24cm. What is the minimum uncertainty implied in this measurement?
Answer
The minimum uncertainty implies in this measurement is $\pm0.01\text{cm}.$
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Question 471 Mark
Calculate the number of nm in $5839\mathring{\text{A}}.$
Answer
$1\text{A} = 10-10\text{m}, 1\text{nm} = 10-9\text{m}$
$\Rightarrow5839\mathring{\text{A}}=583.9\text{nm}.$
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Question 481 Mark
What is effect of temperature on molarity of solution?
Answer
It decreases with increase in temperature because volume of solution increases with increase in temperature.
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