CBSE 11th Standard Chemistry Subject Some Basic Concept of Chemistry Ncert Exemplar 3 Mark Questions 2021
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CBSE 11th Standard Chemistry Subject Some Basic Concept of Chemistry Ncert Exemplar 3 Mark Questions 2021
11th Standard CBSE
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Reg.No. :
Chemistry
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A measured temperature on Fahrenheit scale is 200o F. What will this reading be on Celsius scale?
(a) -
45.4 L of dinitrogen reacted with 22.7 L of dioxygen and 45.4 L nitrous oxide was formed. The reaction is given below
2 N2(g) + O2(g) \(\longrightarrow \) 2 N2O (g)
Which law is being obeyed in this experiment? Write the statement of the law.(a) -
If 4g of NaOH dissolves in 36g of H2O, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1 g mL-1).
(a) -
If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio.
(i) Is this statements true ?
(ii) If yes, according to which law ?
(iii) Give one example related to this law.(a) -
A box contains some identical red coloured balls, labelled as A, each weighing 2 g. Another box contains identical blue coloured balls, labelled as B1 each weighing 5 g. Consider the combinations AB, AB2, A2 B and A2B3 and shows that law of multiple proportion is applicable.
(a)
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CBSE 11th Standard Chemistry Subject Some Basic Concept of Chemistry Ncert Exemplar 3 Mark Questions 2021 Answer Keys
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There are three common scales to measure temperature oC (degree celsius), oF(degree Fahrenheit) and K (kelvin). The K is the SI unit.
The temperature on two scales are related to each other by the following relationship \(^{ 0 }F=\frac { 9 }{ 5 } t^{ 0 }C+32\)
Putting the values in above equation.
\(200-32=\frac { 9 }{ 5 } t^{ 0 }C\Rightarrow \frac { 9 }{ 5 } t^{ 0 }C=168\)
\(\Rightarrow \ t^{ 0 }C=\frac { 168\times 5 }{ 9 } =93.3^{ 0 }C\) -
\( 2 \mathrm{~N}_{2}(g) +\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_{2} \mathrm{O}(\mathrm{g}) \\ 2 \mathrm{~V} \quad \quad\quad 1 \mathrm{~V} \quad \quad \quad 2 \mathrm{~V} \)
45.4 L \(\longrightarrow \) 22.7 L \(\longrightarrow \) 45.4 L
\(\frac { 45.4 }{ 22.7 } = 2\) \(\frac { 22.7 }{ 22.7 } = 1\) \(\frac { 45.4 }{ 22.7 } = 2\)
Hence, the ratio between the volume of the rectants and the product in the given question is simple
i.e. 2: 1 : 2. It proves the Gay Lussac's law of gaseous volumes.
For Gay Lussac's law of gaseous volumes. -
To proceed the calculation, first, calculate the number of moles of NaOH and H2O
Then, find mole fraction of NaOH and H2O by using the
formula \({ { X }_{ NaOH }= }\frac { \eta _{ NaOH } }{ \eta _{ NaOH }+{ \eta }_{ { H }_{ 2 }O } } \left( orX_{ { H }_{ 2 }O }\frac { { \eta }_{ { H }_{ 2 }O } }{ \eta _{ NaOH }+{ \eta }_{ { H }_{ 2 }O } } \right) \)
Then calculate molarity \(=\frac { W\times 1000 }{ m\times V } \) so in order to calculate molarity we reqiure volume of solution which is, \(V=\frac { m }{ specific \ gravity } \)
Number of moles of NaOH, \(\eta _{ NaOH }=\frac { 4 }{ 40 } =0.1mol \ \left\{ \because n=\frac { Mass(g) }{ Molar \ mass(gmol^{ -1 }) } \right\} \)
Similarly, \(\eta _{ { H }_{ 2 }O }=\frac { 36 }{ 18 } =2 \ mol\)
Mole fraction of NaOH,
\(X_{ NaOH }=\frac { moles \ of \ NaOh }{ moles \ of \ NaOH+moles \ of \ { H }_{ 2 }O }\)
\( X_{ NaOH }=\frac { 0.1 }{ 0.1+2 } =0.0476\)
Similarly, \(X_{ { H }_{ 2 }O }=\frac { { \eta }_{ H_{ 2 }O } }{ { \eta }_{ NaOH }+{ \eta }_{ { H }_{ 2 }O } } =\frac { 2 }{ 0.1+2 } =0.9524\)
Total ,mass of solution = mass of solute + mass of solvent
= 4 + 36 = 40 g
Volume of solution
\(=\frac { mass \ of \ solution }{ specific \ gravity } =\frac { 40g }{ 1 \ gm{ L }^{ -1 } } =40 \ mL\)
Similarly,\(=\frac { mole \ of \ solute\times 1000 }{ volume \ of \ solution(mL) } =\frac { 0.1\times 1000 }{ 40 } =2.5M\) -
(i) Yes
(ii) According to the law of multiple proportions.
(iii) \( \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \\ 2 \mathrm{~g} \quad 16 \mathrm{~g} \quad 18 \mathrm{~g} \)
\(\mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2} \\ 2 \mathrm{~g} \quad 32 \mathrm{~g} \quad34 \mathrm{~g} \)
Here, masses of oxygen, (i.e. 16 g in H2 O and 32 g in H2 O2) which combine with fixed mass of hydrogen (2 g) are in the simple ratio i.e. 16 : 32 or 1 : 2. -
Combination Mass of A (g) Mass of B (g) AB 2 5 AB2 2 10 A2B 4 5 A2B3 4 15 Mass of B which is combined with fixed mass of A (say 1 g) will be 2.5 g, 5 g, 1.25 g and 3.75 g in AB, AB2, A2B and A2B3 respectively. They are in the ratio 2 : 4 : 1 : 3. which is simple whole number ratio. Hence, the law of multiple proportion is applicable.
