Question 513 Marks
A flask P contains 0.5 mole of oxygen gas. Another flask Q contains 0.4 mole of ozone gas. Which of the two flasks contain greater number of oxygen atoms?
Answer1 molecule of oxygen $\left(\mathrm{O}_2\right)=2$ atoms of oxygen
1 molecule of ozone $\left(\mathrm{O}_3\right)=3$ atoms of oxygen
In flask P, 1 mole of oxygen gas $=6.022 \times 10^{23}$ molecules
$\therefore 0.5$ mole of oxygen gas $=6.022 \times 10^{23} \times 0.5$ molecules
$=6.022 \times 10^{23} \times 0.5 \times 2$ atoms $=6.022 \times 10^{23}$ atoms
In flask Q 1 mole of ozone gas $=6.022 \times 10^{23}$ molecules
0.4 mole of ozone gas $=6.022 \times 10^{23} \times 0.4$ molecules
$=6.022 \times 10^{23} \times 0.4 \times 3$ atoms $=7.23 \times 10^{23}$ atoms
$\therefore$ Flask $Q$ has greater number of oxygen atoms as compared to the flask $P$.
View full question & answer→Question 523 Marks
A mixture of oxalic acid and formic acid is heated with concentrated $H_2SO_4$ and the gas evolved is collected. On treating the solution with KOH, the volume of the solution decreases by $\frac{1}{6}^\text{th}$ Calculate the molar ratios of two acids in the original mixture.
AnswerLet x mole of oxalic acid and y mole of formic acid be heated with conc. $H_2SO_4$ according to the following equations. $\text{COOH}-\text{COOH}\xrightarrow{\text{H}_2\text{SO}_4/{\text{heat}}}\text{CO(g)}+\text{CO}_2\text{(g)}+\text{H}_2\text{O(l)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x mol} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x mol}\ \ \ \ \ \ \ \text{x mol}$ $\text{HCOOH}\xrightarrow{\text{H}_2\text{SO}_4/{\text{heat}}}\text{CO(g)}+\text{H}_2\text{O(l)}\\\text{y mol}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{y mol}$ Total moles of gaseous mixture = Moles of CO + Moles of $CO_2$ = (x + y) mol + x mol = (2x + y)mol Now, KOH absorbs only Co, i.e. x moles and the volume of the solution decreases by $\frac{1}{6}^\text{th}$ of its volume. Since equal volume of gases have equal number of moles according to Avogadro's law,$\therefore\frac{\text{moles of CO}_2}{\text{moles of both the gases}}$
$=\frac{\text{x}}{(2\text{x}+\text{y})}=\frac{1}{6}$ $\Rightarrow 6\text{x}=2\text{x}+\text{y}$ $\Rightarrow 4\text{x}=\text{y}$ $\Rightarrow \frac{\text{y}}{\text{x}}=4$ $\therefore$ Molar ratio of formic acid : oxalic acid = 4 : 1.
View full question & answer→Question 533 Marks
Describe what you need to do in the laboratory to test (i) the law of conservation of mass, (ii) the law of definite proportion and (iii) the law of multiple proportions.
Answer
- To test the law of conservation of mass, a reaction would have to be carried out in which the mass of the reactants and the mass of the products are weighed and shown to be the same.
- The law of definite proportions could be shown by demonstrating that no matter how a compound is obtained, the reactants remain at the same proportions by mass. This can be done by decomposing a compound and showing that the masses of the elements present are always in the same ratio.
- To test the law of multiple proportions, two different compounds made up of the same elements would have to be decomposed. If the mass of one of the elements is kept constant the masses of other elements combining with that of the element in different samples would have to be in the small whole number ratio.
View full question & answer→Question 543 Marks
- How can you say that sugar is solid and water is liquid?
- Oxygen is prepared by catalytic decomposition of potassium chlorate ($KClO_3$). Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen ($O_2$)· If 2.4mol of oxygen is needed for an experiment, how many grams of potassium chlorate must be decomposed?
(At. mass of K = 39, Cl = 35.5, O = 16)AnswerSugar has fixed shape and volume where as water has fixed volume but not fixed shape.
$2 KClO_3(s) \rightarrow 2 KCl(s)+3 O_2(g)$
Molar mass of $KClO _3=39+35.5+3 \times 16$
$=122.5$
For 3 moles of $O _2$, we need $2 \times 122.5 g$ of. $KClO _3$
For 2.4 moles of $O _2$, we need $=\frac{2 \times 122.5}{3} \times 2.4$
$=196 g \text { of } KClO_3$
View full question & answer→Question 553 Marks
Pressure is determined as force per unit area of the surface. The SI unit of pressure, Pascal is as shown below: $1 \mathrm{~Pa}=1 \mathrm{~N} \mathrm{~m}^{-2}$
If mass of air at sea level is $1034 \mathrm{~g} \mathrm{~cm}^{-2}$, calculate the pressure in Pascal.
AnswerPressure is defined as force acting per unit area of the surface.
$P=\frac{F}{A}$
$=\frac{1034 \mathrm{~g} \times 9.8 \mathrm{~ms}^{-2}}{\mathrm{~cm}^2} \times \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} \times \frac{(100)^2 \mathrm{~cm}^2}{1 \mathrm{~m}^2}$
$=1.01332 \times 10^5 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2}$
We know,
$1 \mathrm{~N}=1 \mathrm{~kg} \mathrm{~ms}^{-2}$
Then,
$1 \mathrm{~Pa}=1 \mathrm{Nm}^{-2}=1 \mathrm{~kg} \mathrm{~m}^{-2} \mathrm{~s}^{-2}$
$1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2}$
$\therefore \text { Pressure }=1.01332 \times 10^5 \mathrm{~Pa}$
View full question & answer→Question 563 Marks
Two oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is $\mathrm{M}_3 \mathrm{O}_4$, find that of the second.
AnswerIn the first oxide, oxygen $=27.6$
Metal $=100-27.6=72.4$ parts by mass.
As the formula of the oxide is $\mathrm{M}_3 \mathrm{O}_4$ it means
72.4 parts by mass of metal $=3$ atoms of metal and
4 atoms of oxygen $=27.6$ parts by mass.
In the second oxide, oxygen $=30.0$ parts by mass and metal $=100-30=70$ parts by mass.
But 72.4 parts by mass of metal $=3$ atoms of metal.
$\therefore 70$ parts by mass of metal $=\frac{3}{72.4} \times 70$ atoms of metal $=2.90$ atoms of metal
Also, 27.6 part by mass of oxygen $=4$ atoms of oxygen.
$\therefore 30$ part by mass of oxygen $=\frac{4}{27.6} \times 30$ atoms of oxygen
$=4.35$ atoms of oxygen.
Hence, ratio of $\mathrm{M}: \mathrm{O}$ in the second oxide $=2.90: 4.35=1: 1.5$ or $2: 3$
$\therefore$ Formula of the other metal oxide is $\mathrm{M}_2 \mathrm{O}_3$.
View full question & answer→Question 573 Marks
Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
AnswerMole fraction of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ $=\frac{\text{Number of moles of }\text{C}_2\text{H}_5\text{OH}}{\text{Number of moles of solution}}$
$0.040=\frac{^\text{n}\text{C}_2\text{H}_5\text{OH}}{^\text{n}\text{C}_2\text{H}_5\text{OH}+\ ^\text{n}\text{H}_2\text{O}}\ .....(1)$
Number of moles present in 1L water:
$^\text{n}\text{H}_2\text{O}=\frac{1000\text{g}}{18\text{g }\text{mol}^{-1}}$
$^\text{n}\text{H}_2\text{O}=55.55\text{mol}$
Substituting the value of $^\text{n}\text{H}_2\text{O}$ in equation (1),
$\frac{^\text{n}\text{C}_2\text{H}_5\text{OH}}{^\text{n}\text{C}_2\text{H}_5\text{OH}+55.55}=0.040$
$^\text{n}\text{C}_2\text{H}_5\text{OH}=0.040\ ^\text{n}\text{C}_2\text{H}_5\text{OH}+(0.040)(55.55)$
$0.96\ ^\text{n}\text{C}_2\text{H}_5\text{OH}=2.222\text{mol}$
$^\text{n}\text{C}_2\text{H}_5\text{OH}=\frac{2.222}{0.96}\text{mol}$
$^\text{n}\text{C}_2\text{H}_5\text{OH}=2.314\text{mol}$
$\therefore$ Molarity of solution $=\frac{2.314\text{mol}}{1\text{L}}$
$= 2.314 \text{M}$
View full question & answer→Question 583 Marks
Cone. HCl is $38 \% \mathrm{~HCl}$ by mass. What is the molarity of this solution if $\mathrm{d}=1.19 \mathrm{~g} \mathrm{~cm}^{-3}$ ? What volume of cone. HCl is required to make 1.00 L of 0.10 M HCl ?
Answer$\text{M}=\frac{\text{W}_{\text{B}}}{\text{M}_{\text{B}}}\times\frac{1000}{\text{Vol. of solution in ml}}$
$=\frac{38}{36.5}\times\frac{100}{\frac{\text{Mass of solution}}{\text{Density of solution}}}$
$=\frac{38}{36.5}\times\frac{1000}{\frac{100}{1.19}}$
$=\frac{38}{36.5}\times\frac{1000\times1.19}{100}=12.39\text{M}$
$\text{M}_1\text{V}_1=\text{M}_2\text{V}_2$
$\Rightarrow12.39\times\text{V}_1=0.1\times100\text{ml}$
$\Rightarrow\text{V}_1=\frac{0.1\times1000}{12.39}\text{M}$
$\Rightarrow\text{V}_1=8.07\text{ml}.$
View full question & answer→Question 593 Marks
A compound on analysis was found to contain C = 34.6%, H = 3.85%, and 0 = 61.55%. Calculate its empirical formula.
AnswerCalculation of the empirical formula.
| Elemenl |
Percentage |
Atomic mass |
Gram atoms (Moles) |
Molar ratio |
Simplest whole numbar ratio |
| C |
34.6 |
12 |
$\frac{34.6}{12}=2.88$ |
$\frac{2.88}{2.88}=1$ |
3 |
| H |
3.85 |
1 |
$\frac{3.85}{1}=3.85$ |
$\frac{3.85}{2.88}=1.335\text{ or }\frac{4}{3}$ |
4 |
| O |
61.65 |
16 |
$\frac{61.55}{16}=3.85$ |
$\frac{3.85}{2.88}=1.335\text{ or }\frac{4}{3}$ |
4 |
$\therefore$ The simplest whole number ratios of the different elements are: C : H : 0 :: 3 : 4 : 4 and the empirical formula of the compound = $\mathrm{C}_3 \mathrm{H}_4 \mathrm{O}_4$. View full question & answer→Question 603 Marks
If the density of methanol is $0.793 \mathrm{~kg} \mathrm{~L}^{-1}$, what is its volume needed for making 2.5 L of its 0.25 M solution?
AnswerMolar mass of methanol $\left(\mathrm{CH}_3 \mathrm{OH}\right)=(1 \times 12)+(4 \times 1)+(1 \times 16)$
$=32 \mathrm{~g} \mathrm{~mol}^{-1}$
$=0.032 \mathrm{~kg} \mathrm{~mol}^{-1}$
Molarity of methanol solution $=\frac{0.793 \mathrm{~kg} \mathrm{~L}^{-1}}{0.032 \mathrm{~kg} \mathrm{~mol}^{-1}}$
$=24.78 \mathrm{~mol} \mathrm{~L}^{-1}$
(Since density is mass per unit volume)
Applying,
$M_1 V_1=M_2 V_2$
(Given solution) (Solution to be prepared)
$\left(24.78 \mathrm{~mol} \mathrm{~L}^{-1}\right) \mathrm{V}_1=(2.5 \mathrm{~L})\left(0.25 \mathrm{~mol} \mathrm{~L}^{-1}\right)$
$\mathrm{V}_1=0.0252 \mathrm{~L}$
$\mathrm{~V}_1=25.22 \mathrm{~mL}$
View full question & answer→Question 613 Marks
A crystalline salt when heated becomes anhydrous and loses 51.2% of its weight. The anhydrous salt on analysis gave the percentage composition as Mg = 20.0%, S = 26.66% and O = 53.33%.
Answer
| Element |
% of mass |
Atomic mass |
Relative no. of moles element |
Simple molar ratio |
| Mg |
20 |
24 |
$\frac{20}{24}=0.8333$ |
$\frac{0.8333}{0.8333}=1$ |
| S |
26.66 |
32 |
$\frac{26.66}{32}=0.8125$ |
$\frac{0.8125}{0.8333}=1$ |
| O |
53.33 |
16 |
$\frac{53.33}{16}=3.333125$ |
$\frac{3.333}{0.8333}=4$ |
The empirical formula of the anhydrous salt comes out to be $\mathrm{MgSO}_4$. Empirical formula mass $=120$.
Molecular mass $=120$.
Hence, molecular formula $=\mathrm{MgSO}_4$.
As crystalline salt on becoming anhydrous loses $51.2 \%$ by mass, this means 48.8 g of anhydrous salt contains $\mathrm{H}_2 \mathrm{O}=$
51.2 g . Therefore, 120 g of anhydrous salt contains
$=\frac{51.2}{48.8} \times 120 \mathrm{~g}=126 \mathrm{~g} \frac{126}{18}$ molecules $=7 \mathrm{H}_2 \mathrm{O}$ molecules
Hence, molecular formula of crystalline salt $=\mathrm{MgSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O}$ View full question & answer→Question 623 Marks
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, $1.41 g mL ^{-1}$ and the mass per cent of nitric acid in it being $69 \%$.
AnswerMass percent of nitric acid in the sample = 69% [Given]
Thus, 100g of nitric acid contains 69g of nitric acid by mass.
Molar mass of nitric acid ($HNO_3$).
$= {1 + 14 + 3(16)}g mol^{–1}$
$= 1 + 14 + 48$
$= 63g mol^{–1}$
$\therefore$ Number of moles in 69g of $HNO_3.$
$=\frac{69\text{g}}{63\text{g mol}^{-1}}$
$=1.095\text{ mol}$
Volume of 100g of nitric acid solution,
$=\frac{\text{Mass of solution}}{\text{density of solution}}$
$=\frac{100\text{g}}{1.41\text{g mL}^{-1}}$
$=70.92\text{mL}=70.92\times10^{-3}\text{L}$
Concentration of nitric acid,
$=\frac{1.095\text{mole}}{70.92\times10^{-3}\text{L}}$
$=15.44\text{mol/L}$
$\therefore$ Concentration of nitric acid = 15.44mol/L.
View full question & answer→Question 633 Marks
How much copper can be obtained from 100g of copper sulphate ($\mathrm{CuSO}_4$)?
Answer1 mole of $\mathrm{CuSO}_4$ contains 1 mole of Cu .
Molar mass of $\mathrm{CuSO}_4$
$=(63.5)+(32.00)+4(16.00)$
$=63.5+32.00+64.00$
$=159.5 \text { gram }$
159.5 gram of $\mathrm{CuSO}_4$ contains 63.5 gram of Cu .
Therefore, 100g of $\mathrm{CuSO}_4$ will comtain $\frac{63.5\times100\text{g}}{159.5}$ of Cu.
$=\frac{63.5\times100\text{g}}{159.5}$
$=39.81\text{g}$
View full question & answer→Question 643 Marks
How many grams of $Cl _2$, are required to completely react with 0.4 g of H, to yield HCl ? Also, calculate the amount of HCl formed.
Answer$\text{H}_2\text{(g)} \ \ \ \ \ \ \ \ +\ \ \ \ \ \ \ \text{Cl}_2\text{(g)} \xrightarrow{\ \ \ \ \ \ \ } 2\text{HCl(g)}\$2\times1=2)\ \ \ \ \ (2\times35.5=71)$
Now, 2 g of $H _2$ reacts with 71 g of $Cl _2$ to give 73 g of HCl .
So, 0.4 g of $H _2$ reacts with $\frac{71}{2} \times 0.4=14.2 g$ of $Cl _2$ to given
$\frac{73 \times 0.4}{2}=14.6 g$ of HCl
Therefore, 14.2 g of $Cl _2$ and 14.6 g of HCl are required.
View full question & answer→Question 653 Marks
Hydrogen reacts with nitrogen to produce ammonia according to the equation:
$3H_2(g) + N_2(g) → 2NH_3(g)$
Determine How much ammonia would be produced if 100g of $N_2$ reacts?
Answer$3\text{H}_2\text{(g)}\ \ \ \ +\ \ \ \ \ \text{N}_2\text{(g)}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }2\text{NH}_3\text{(g)}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2\times14=28\text{g})\ \ \ (2\times17=34\text{g})$
Now, 28g of $N_2$, reacts with hydrogen to form 34g of $NH_3$
Hence, 100g of $N_2$ reacts with hydrogen to form
$\frac{34}{28}\times100=\frac{3400}{28}=121.4\text{g}.$
View full question & answer→Question 663 Marks
Calculate:
- Mass in grams of 5.8mol of $N_2O.$
- Number of moles in 8.0g of $O_2.$
- Molar mass of 11.2L at STP weighs 8.5g.
Answer1 mole of $H_2O = 28 + 16 = 44g$
$5.8$ moles of $H_2O = 44 \times 5.8g = 255.2g$
Number of moles $=\frac{\text{Given mass}}{\text{Molar mass of O}_2}$
$=\frac{8.0\text{g}}{32\text{g}}=0.25\text{ mole.}$
11.2L of $NH_2$ at STP weighs $= 8.5g$
$22.4 L$ of $NH_2$ at STP weighs $= 17g$
Molar mass $= 17g mol^{-1}$
[$\because$ 1 mol of every gas at STP has V = 22.4 at STP]
View full question & answer→Question 673 Marks
What is the concentration of sugar $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ in $\mathrm{mol}^{-1}$ if its 20 g are dissolved in enough water to make a final volume up to 2L?
AnswerMolarity (M) of a solution is given by,
$=\frac{\text{Number of moles of solute}}{\text{Volume of solution in Litres}}$
$=\frac{\text{Mass of suger/molar mass of sugar}}{2\text{L}}$
$=\frac{20\text{g}/[(12\times12)+(1\times22)+(11\times16)]\text{g}}{2\text{L}}$
$=\frac{20\text{g}/342\text{g}}{2\text{L}}$
$=\frac{0.0585\text{ mol}}{2\text{L}}$
$= 0.02925 \text{ mol L}^{–1}$
$\therefore$ Molar concentration of sugar = 0.02925 mol $\text{L}^{–1}$
View full question & answer→Question 683 Marks
Air contains $20\%$ oxygen by volume. Calculate the theoretical volume of air which will be required for burning completely $500\ m$ of acetylene gas.All volumes are measured under the same conditions of temperature and pressure.
Answer
| $2C_2H_2$ |
$+5O_2$ |
$→4CO_2$ |
$+2H_2O(g)$ |
| 2vol |
5vol |
4vol |
2vol (Gay-Lussac's Law) |
| Or 1vol |
$\frac{5}{2}\text{vol}$ |
$\frac{4}{2}\text{vol}$ |
$\frac{2}{2}\text{vol}$ |
| $500m^3$ |
$\frac{5}{2}\times500\text{cm}^3$ |
$\frac{4}{2}\times500\text{cm}^3$ |
$\frac{2}{2}\times500\text{cm}^3$ |
| $500m^3$ |
$1250m^3$ |
$1000m^3$ |
$500m^3$ |
Thus, $1250m^3$ oxygen is required for burning $500m^3$ of acetylene. But the percentage of oxygen in air is 20%
$\therefore$ Volume of air required $= 1250 \times \frac{100}{20} = 6250\text{m}^3.$ View full question & answer→Question 693 Marks
Calculate the mass percent of different elements present in sodium sulphate ($\mathrm{Na}_2 \mathrm{SO}_4$).
AnswerThe molecular formula of sodium sulphate is $\mathrm{Na}_2 \mathrm{SO}_4$. Molar mass of $\mathrm{Na}_2 \mathrm{SO}_4=[(2 \times 23.0)+(32.066)+4(16.00)]$
$=142.066 \mathrm{~g}$
Mass percent of an element $=\frac{\text{Mass of that element in the compound}}{\text{Molar mass of compoud}}\times100$
$\therefore$ Mass percent of sodium:
$=\frac{46.0\text{g}}{142.066\text{g}}\times100$
$=32.379$
$=32.4\%$
Mass percent of sulphur:
$=\frac{32.066\text{g}}{142.066\text{g}}\times100$
$=22.57$
$=22.6\%$
Mass percent of oxygen:
$=\frac{64.0\text{g}}{142.066\text{g}}\times100$
$=45.049$
$=45.05\%$
View full question & answer→Question 703 Marks
Calculate the atomic mass (average) of chlorine using the following data:
|
|
% Natural Abundance
|
Molar Mass
|
| ${ }^{35} \mathrm{Cl}$ |
75.77
|
34.9689
|
| ${ }^{35} \mathrm{Cl}$ |
24.23
|
36.9659
|
AnswerThe average atomic mass of chlorine
$=\begin{bmatrix}\begin{pmatrix}\text{Fractional abundance}\\\text{ of }\ ^{35}\text{Cl}\end{pmatrix}\begin{pmatrix}\text{Molar mass}\\\text{of }\ ^{35}\text{Cl} \end{pmatrix}\\+\begin{pmatrix}\text{Fractional abundance}\\\text{ of }\ ^{35}\text{Cl}\end{pmatrix}\begin{pmatrix}\text{Molar mass}\\\text{of }\ ^{35}\text{Cl} \end{pmatrix}\end{bmatrix}$
$=\Big[\Big\{\Big(\frac{75.77}{100}\Big)(34.9689)\Big\}+=\Big[\Big\{\Big(\frac{24.23}{100}\Big)(36.9659)\Big\}\Big]$
$= 26.4959 + 8.9568$
$= 35.4527\text{u}$
$\therefore$ The average atomic mass of chlorine = 35.4527u.
View full question & answer→Question 713 Marks
Match Cloumn I with Column II.
|
S. No
|
Cloumn I
|
S. No
|
Cloumn II
|
|
1.
|
88g of $CO_2$
|
(i)
|
0.25mol
|
|
2.
|
$6.022 \times 10^{23}$ molecules of $H_2O$
|
(ii)
|
2mol
|
|
3.
|
5.6 litres of $O_2$ at STP
|
(iii)
|
1mol
|
|
4.
|
96g of $O_2$
|
(iv)
|
$6.022 \times 10^{23}$ molecules
|
|
5.
|
1mol of any gas
|
(v)
|
3mol
|
Answer
|
S. No
|
Cloumn I
|
S. No
|
Cloumn II
|
|
1.
|
88g of $CO_2$
|
(ii)
|
2mol
|
|
2.
|
$6.022 \times 10^{23}$ molecules of $H_2O$
|
(iii)
|
1mol
|
|
3.
|
5.6 litres of $O_2$ at STP
|
(i)
|
0.25mol
|
|
4.
|
96g of $O_2$
|
(v)
|
3mol
|
|
5.
|
1mol of any gas
|
(iv)
|
$6.022 \times 10^{23}$ molecules
|
View full question & answer→Question 723 Marks
A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical blue coloured balls, labelled as B, each weighing 5 grams. Consider the combinations $AB, AB_{2,} A_2B$ and $A_2B_3$ and show that law of multiple proportions is applicable.
AnswerAccording to the law of multiple pro portion, when two elements comvune to from two or more compoundds, then the different massses one element, which combine with a fixed mass of the other, bear a simple ratio to one another.
1g of A combines with $\frac{5}{2}\text{g}$ of B = 2.5g B
For $AB_2$
1g of A combines with $\frac{10}{2}\text{g}$ of B = 5g of B
For $A_2B$
1g of A combines wuth $\frac{5}{4}\text{g}$ of B = 1.25g of B
For $A_2B_2$
1g of A combines with $\frac{15}{4}\text{g}$ of B = 3.75g of B
Thus, it is proved thed law mukltiple proportions is applicable.
View full question & answer→Question 733 Marks
Calculate the number of moles in the following masses,
- 1.46 metric ton of Al (1 metric ton $= 10^3kg).$
- 7.9 mg of Ca.
Answer
- 1.46 metric ton of Al $= 1.46 \times 10^3 \times 10^3g of Al$
$= 1.46 \times 10^6g$
Atomic mass of Al = 27
Moles of Al $=\frac{\text{mass of Al}}{\text{atomic mass}}=\frac{1.46\times10^{46}}{27}$
$= 5.41 \times 10^4mol$
- 7.9mg of Ca $= 7.9 \times 10^{-3}g of Ca$
Atomic mass of Ca = 40.1
mole of Ca $=\frac{\text{mass of Ca}}{\text{atomic mass}}=\frac{7.9\times106{-3}}{40.1}$
$=1.97\times10^{-4}\text{mol}.$ View full question & answer→Question 743 Marks
For an actual result of an observation to be 5; two students A and B reported their readings as follows:
|
|
Observation number
|
Average
|
|
1
|
2
|
|
Student A
|
4.95
|
4.93
|
4.94
|
|
Student B
|
4.94
|
5.05
|
4.995
|
Which of the students has made a more precise observation? Is his observation accurate too?
AnswerStudent 'A' has made a more precise observation since the variation in the two readings taken by him is not much.
His observation is precise but is not accurate since his readings are not close to the actual reading which is 5. Student B is more accurate as his average reading is close to actual reading i.e. 5.
View full question & answer→Question 753 Marks
Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.
Note: Given that the molar mass of the oxide is $159.69 \mathrm{~g} \mathrm{~mol}^{-1}$
Answer$\%$ of iron by mass $=69.9 \%$ [Given]
$\%$ of oxygen by mass $=30.1 \%[$ Given]
Atomic mass of iron $=55.85 \mathrm{amu}$.
Atomic mass of oxygen $=16.00 \mathrm{amu}$.
Relative moles of iron in iron oxide $=\%$ mass of iron by mass/Atomic mass of iron $=69.9 / 55.85=1.25$
Relative moles of oxygen in iron oxide $=\%$ mass of oxygen by mass/Atomic mass of oxygen $=30.01 / 16=1.88$
Simplest molar ratio $=1.25 / 1.25: 1.88 / 1.25$
$\Rightarrow 1: 1.5=2: 3$
$\therefore$ The empirical formula of the iron oxide is $\mathrm{Fe}_2 \mathrm{O}_3$.
Mass of $\mathrm{Fe}_2 \mathrm{O}_3=(2 \times 55.85)+(3 \times 16.00)=159.7 \mathrm{~g} \mathrm{~mol}^{-1} \mathrm{n}=$ Molar mass $/$ Empirical formula mass $=159.7 / 159.6=$ 1(approx)
Thus, Molecular formula is same as Empirical Formula i.e. $\mathrm{Fe}_2 \mathrm{O}_3$.
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QUESIION Potassium superoxide, $KO _2$ is used in rebreathing gas masses to generate oxygen
$4 KO_2(s)+2 H_2 O(l) \rightarrow 4 KOH(s)+3 O_2(g)$
If a reaction vessel contains $0.15{mol KO _2}^2$ and $0.10 mol H _2 O$, what is limiting reactant. Bow many moles of oxygen can be produced?
Answer$4$ moles of $KO _2$ reacts with 2 moles of $H _2 O$
$0.15$ mole of $KO _2$ will react with $\frac{2}{4} \times 0.15=0.075$ moles of water.
But we have 0.10 mol of $H _2 O$
$\therefore H _2 O$ is excess reagent, $KO _2$ is limiting reagent.
$4$ moles of $KO _2$ gives 3 moles of $O _2$
$0.15$ mole of $KO _2$ will given $\frac{3}{4} \times 0.15=\frac{0.45}{4}=0.1125$ moles of $O _2$.
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A 0.005 cm thick coating of copper is deposited on a plate of $0.5 \mathrm{~m}^2$ total area. Calculate the number of copper atoms deposited on the plate 0 (density of copper $=7.2 \mathrm{~g} \mathrm{~cm}^{-3}$, atomic mass $=63.5$ ).
AnswerArea of plate $=0.5 \mathrm{~m}^2=0.5 \times 10^4 \mathrm{~cm}^2$
Thickness of coating $=0.005 \mathrm{~cm}$
Volume of copper deposited $=0.5 \times 10^4 \times 0.005=25 \mathrm{~cm}^3$
Mass of copper deposited $=25 \times 7.2=180 \mathrm{~g}$
Now, 63.5 g of copper contains atoms $=6.022 \times 10^{23}$
$\therefore 180 \mathrm{~g}$ of copper will contain atoms $=\frac{6.022 \times 10^{23}}{36.5} \times 180$
$=1.71 \times 10^{23}$ atoms.
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