NCERT Solutions for Class 11 Chemistry
Chapter 1- Some Basic Concepts of Chemistry
Q1. Calculate the molar mass of the following: (i) H2O(ii) CO2 (iii) CH4
Ans.
(i) Molar mass of water is calculated as below:
Mass of one mole of H atoms = 1g/mol
Mass of one mole of O atoms = 16g/mol
Since we have two H atoms and one O atom in H2O, hence the mass of one mole of H2O will be = (2×1+16)g/mol = 18g/mol
(ii) Molar mass of Carbon Dioxide is calculated as below:
Mass of one mole of C atoms = 12g/mol
Mass of one mole of O atoms = 16g/mol
Since we have one atom of C and two atoms of O, hence the molar mass of CO2
will be = (12+16×2) = 44g/mol
(iii)Molar mass of Methane is calculated as below:
Mass of one mole of C atoms = 12g/mol
Mass of one mole of H atoms = 1g/mol
Since we have one atom of Cand four atoms of H in CH4, hence the molar mass of
CH4will be (12+1×4)= 16g/mol
Q2. Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4).
Ans.
Molar mass of
Na2SO4= [(2 x 23.0) + (32.066) + 4(16.00)]
=142.066 g/mol
Formula to calculate mass percent of an element =
(Mass of element in the compound /total mass of compound) x 100
Therefore, mass percent of the sodium will be:
=(46.0g /142.066g)×100
= 32.38= 32.4%
Mass percent of the sulphur elementwill be:
=(32.066g/142.066g)×100
= 22.6= 22.6%
Mass percent of the oxygen elementwill be:
=(64.0g/142.066g)×100
= 45.049= 45.05%
Q3. Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.
Ans.
We know that atomic mass of iron is 55.85u and percentage of iron given is 69.9%.
Hence, the relative moles for oxygen in iron oxide will be :
(Percentage of iron / atomic mass of iron)
(69.9/55.85)
= 1.25
Also, the relative moles for oxygenin iron oxide will be :
(Percentage of oxygen/ atomic mass ofoxygen)
(30.1/16)
= 1.88
Hence, the molar ratio of iron to oxygen will be obtained as:
⇒ 1.25: 1.88 ⇒1: 1.5 ⇒2: 3
Hence, the empirical formula of the iron oxide is Fe2O3
.
Q4. Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Ans.
(i) 1 mole of carbon is burnt in air.
The equation for this will be as:
C + O2⇒CO2
From equation we get that 1 mole of carbon reacts with 1 mole of O2 to form one mole of CO2. Hence, amount of CO2 produced = Molar mass of CO2= 44 g
(ii)1 mole of carbon is burnt in 16 g of O2.
From above equation we get that1 mole of carbon reacts with 1 mole (32g) of O2 to form 44g of CO2. Hence the amount of CO2 produced when 1 mole of carbon is burnt in 16 g of dioxygen = (44×16)/32 = 22g
(iii) 2 moles of carbon are burnt in 16 g of O2.
Since 16 g of O2 gives 22g of CO2, and also here, the O2 is the limiting reactant. So in this case too, CO2 produced will be 22g.
Q5.Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1.
Ans.
0.375 M aqueous solution of CH3COONa means that 0.375 moles of CH3COONa is present in 1000 mL of solution.
Hence the number of moles of CH3COONa present in 500 ml of soln. will be
(0.375×500)/1000
= 0.1875 mole
Now it is given that molar mass of CH3COONa is 82.0245g/mol which means that mass of one mol of CH3COONa = 82.0245g
Thus, mass of 0.1875 mol of CH3COONa= (82.02×0.1875) g= 15.378g
Q6. 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%
Ans.
In this question, firstly we will calculate the Molar mass of HNO3.
Molar mass of HNO3 = {1 + 14 + 3(16)} g/mol = 63g/mol.
Mass of given sample = volume×density = 1.41g
Now, percentage of nitric acid present = 69%
So, mass of nitric acid present in sample = (1.41×69)/100 =0.973g
Now since, 63g is the mass of 1 mol of NHO3
Therefore, 0.973g will be mass of (0.973g/63g)mol = 0.01544 mol of HNO3.
Now, 1 ml of soln. contains 0.01544 mol of HNO3
Thus, 1000ml(1L) of soln. contains = (1000*0.01544)mol of HNO3.
Hence, concentration of nitric acid in moles per litre= 15.44mol/L.
Q7. How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
Ans.
From the chemical formula of CuSO4, we see that one CuSO4molecule has one Cu atom.
Hence, one mole of CuSO4will contain 1 mole of Cu atoms.
Now we will calculate the molar mass of CuSO4.
Molar mass of CuSO4= {63.5 + 32.00 + 4(16.00)}g/mol = 159.5 g/mol
So, 159.5 g of CuSO4 contains 63.5g of Cu.
Thus, 100g of CuSO4 will contain {(100*63.5)/159.5}g = 39.81g
Q8. Determine the molecular formula of an oxide of iron, in which the mass percent of iron and oxygen are 69.9 and 30.1, respectively.
Ans.
Here,
Mass percent of Fe = 69.9%
Mass percent of O = 30.1%
Since 55.85g is the mass of 1 mole of Fe.
Thus, 69.9g will be the mass of (69.9/55.85)g = 1.25g of Fe
Similarly,
No. of moles of O present in oxide=(30.1/16.0)g =1.88g of O
So, we get the ratio of Fe to O in oxide as,
= 1.25: 1.88
=1:1.5
=2:3
Therefore, the empirical formula of oxide isFe2O3
Also, the Empirical formula mass ofFe2O3= {2(55.85) + 3(16.00)} g= 159.7g
And the molar mass of Fe2O3= 159.69g
Therefore n = (Empirical formula mass/ molar mass) = 0.999≈ 1
The molecular formula = n * empirical formula.
Thus, the empirical of the given oxide isFe2O3and n obtained is 1.
Thus, the molecular formula of the oxide is Fe2O3
Q9. Calculate the atomic mass (average) of chlorine using the following data:
| Percentage Natural Abundance | Molar Mass | |
| 35Cl | 75.77 | 34.9689 |
| 37Cl | 24.23 | 36.9659 |
Ans.
Abundance of 35Cl = 75.77% = 0.7577 and Molar mass = 34.9689g/mol
Since molar mass is numerically equal to atomic mass,so,
Atomic massof 35Cl = 34.9689u.
Abundance of 37Cl = 24.23% = 0.2423 and Molar mass = 36.9659g/mol
Atomic mass of 37Cl = 36.9659u.
Average Atomic mass can be calculated as = {(0.7577 x 34.9689) + (0.2423 x 36.9659)}u= 26.4959 + 8.9568 = 35.4527u
Q10. In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atom
(iii) Number of molecules of ethane
Ans.
(i) Since we can see that in one molecule of ethane(C2H6), two carbon atoms are present. So, in one mole of ethane, two moles of carbon atoms will be present.
Therefore, in three moles of ethane, (3×2) = 6 moles of carbon atoms will be present.
(ii)In one molecule of ethane(C2H6), six Hydrogen atoms are present. So, in one mole of ethane, six moles of Hydrogen atoms will be present.
Therefore, in three moles of ethane, (3×6) = 18 moles of Hydrogen atoms will be present.
(iii) In one mole of ethane, number of molecules present = 6.023x 1023molecules
Hence, in three moles of ethane, number of molecules present = 3×6.023x 1023
= 18.069 x 1023 molecules.
Q11. What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2L?
Ans.
In this question, first we will calculate the moles of solute.
Molar mass of sugar(C12H22O11) = 342.2965g/mol
So, no. of moles of sugar = (20g/342g) = 0.0585 mol
Thus, molarity is given as (no. of moles of solute)/ (Volume of Soln. in Litres)
Therefore, Molarity = (0.0585/2)mol/L = 0.02925 mol/L
Q12. If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M solution?
Ans.)
Mass of methanol required is
mass=Volume of solution×Molarity of the solution×Molar mass of methanol
=2.5×0.25×32=20 g.
Therefore,
Volume of methanol needed =mass/density=20/0.793=25.22 ml.
Q13. Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
1Pa = 1N m–2
If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal
Ans.
Pressure is the force acting per unit area of the surface.
But weight = mg
∴ Pressure = Weight per unit area
Pressure = weight per unit area =mg
=1034g/cm2×(1kg/1000g)×(100cm2/1m)×9.80m/s2×(1N/kgm/s2)×(1Pa /1N/m2)
=1.013×105Pa
Q14. What is the SI unit of mass? How is it defined?
Ans.
The SI unit of mass is kilogram (kg). It has been defined as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an airtight jar at International Bureau of Weights and Measures in Sevres, France.
Q15. Match the following prefixes with their multiples:
| Prefixes | Multiples | |
| (a) | Femto | 10 |
| (b) | Giga | 10−15 |
| (c) | Mega | 10−6 |
| (d) | Deca | 109 |
| (e) | Micro | 106 |
Ans.
| Prefixes | Multiples | |
| (a) | femto | 10−15 |
| (b) | giga | 109 |
| (c) | mega | 106 |
| (d) | deca | 10 |
| (e) | micro | 10−6 |
Q16. What do you mean by significant figures?
Ans.
Significant figures are meaningful digits which are known with certainty plus one which is estimated or uncertain.
For e.g.: Suppose you obtain a result of the experiment as75.3 ml. So, in this case 75is certain and 3 is uncertainand the uncertainty would be +/-(1) in the last digit(3 in this case). So, the total significant figures are 3.
Q17. A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in per cent by mass.
(ii) Determine the molality of chloroform in the water sample.
Ans.
(i) 1 ppm is defined as 1 part out of 1 million parts.
Mass percent of 15 ppm chloroform in H2O
=(15/106) ×100
≈1.5 ×10−3%
(ii) Molar mass of CHCl3 = 119.38 g/mol
So, moles of CHCl3 present in sample = (15/119.38)mol = 0.125 mol
Therefore, molality = 0.125 mol/ 1000g = 1.25 × 10-4 mol/kg
Q18. Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Ans.
Scientific notation is a way of writing numbers that are either very large or very small. A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10.
(i) 0.0048= 4.8 ×10−3
(ii) 234,000 = 2.34 ×105
(iii) 8008= 8.008 ×103
(iv) 500.0 = 5.000 ×102
(v) 6.0012 = 6.0012 ×100
Q19. How many significant figures are present in the following?
(a) 0.0025
(b) 208
(c) 5005
(d) 126,000
(e) 500.0
(f) 2.0034
Ans.
(a) 0.0025: 2 significant numbers (zeros to the left of 1st non-zero digit are insignificant).
(b) 208: 3 significant numbers (zeros between two non-zero digits are significant).
(c) 5005: 4 significant numbers(zeros between two non-zero digits are significant).
(d) 126,000:3 significant numbers(zeros to the left of decimal and right of non-zero digit are insignificant).
(e) 500.0: 4 significant numbers(zeros to the right of decimal are significant and also, non-zero digit to the left of decimal is significant. So both the middle zeros are present between significant numbers.)
(f) 2.0034: 5 significant numbers(same explanation as above)
Q20. Round up the following upto three significant figures:
(a) 34.216
(b) 10.4107
(c)0.04597
(d)2808
Ans.
(a) The number we will get after round up: 34.2
(b) The number we will get after round up: 10.4
(c)The number we will get after round up: 0.046
(d)The number we will get after round up: 281
Q21. The following data are obtained when dinitrogen and dioxygen react together to form different compounds:
| Mass of dioxygen | Mass of dinitrogen | |
| (i) | 16 g | 14 g |
| (ii) | 32 g | 14 g |
| (iii) | 32 g | 28 g |
| (iv) | 80 g | 28 g |
(a) Which law of chemical combination is obeyed by the above experimental data?
Give its statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = …………………. mm = …………………. pm
(ii) 1 mg = …………………. kg = …………………. ng
(iii) 1 mL = …………………. L = …………………. dm3
Ans.
(a)
Upon fixing the mass of dinitrogen at 14g, the masses of dioxygen that are going to combine with the given fixed mass of dinitrogen are 16g, 32g, 32g, and 80g.
The masses of dioxygen are in a whole number ratio of 1:2:2:5.
Therefore, we can say that the given experimental data obeys the Law of Multiple Proportions.
(b)
(i) 1 km = 106mm = 1015 pm
(ii) 1 mg = 10-6kg = 106 ng
(iii) 1 mL = 10-3 L =10-3 dm3
Q22. If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in 2.00 ns.
Ans.
The given time is in ns, so firstly we will convert it in s.
Time taken = 2 ns= 2 ×10−9s
Now,
Speed of light = 3 ×108 m/s
Now,
Distance = Speed x Time
So,
Distance travelled in 2 ns = speed of light x time taken by light
= (3 ×108)(2 ×10−9) m
= 6 ×10-1m
= 0.6 m
Q23. In a reaction
A + B2 → AB2
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Ans.
Limiting reagentis the first to get fully consumed during a reaction, thus causes the reaction to stop and limits the amount of product formed.It determines the extent of a reaction.
(i) 300 atoms of A + 200 molecules of B
1 atom of A reacts with 1 molecule of B.
So, 200 atoms of A reacts with 200 molecules of B,
Therefore 100 atoms of A remain unconsumed and B is consumed fully.
Hence, B is the limiting reagent.
(ii) 2 mol A + 3 mol B
1 mole of A reacts with 1 mole of B.
So, 2 moles of A reacts with 2 moles of B, so 1 mole of B is unused whereas A is consumed fully.
Hence, A is the limiting reagent.
(iii) 100 atoms of A + 100 molecules of B
1 atom of A reacts with 1 molecule of B.
So, 100 atoms of A will react with 100 molecules of B. Hence there will be no limiting reagent as it is a stoichiometric mixture.
(iv) 5 mol A + 2.5 mol B
1 mole of A reacts with 1 mole of B.
So, 2.5 moles of A reacts with 2.5 moles of B, so 2.5 moles of A is unused whereas B is consumed fully.
Hence, B is the limiting reagent.
(v) 2.5 mol A + 5 mol B
1 mole of A reacts with 1 mole of B.
So, 2.5 moles of A reacts with 2.5 moles of B, so 2.5 moles of B is unused whereas A is consumed fully.
Hence, A is the limiting reagent.
Q24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
N2 (g) + H2(g)→ 2NH3 (g)
(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass.
Ans.
(i) From a balanced equation, we have,
N2 (g) +3H2(g)→ 2NH3 (g)
Hence, we can say that 1 mol(28g) of N2 reacts with 3 mol(6g) of H2.
∴ 2000 g of N2 will react with (6g × 2000g)/28g = 428.6g of H2
Now, 28g of N2 produces 34g of NH3
Hence, 2000g of N2 produces (34 ×2000) / 28 = 2428.57g of NH3
Hence, here, N2 will get consumed fully and H2will remain unreacted.
(ii) H2 will remain unreacted
(iii) Mass of H2 left unreacted = 1000g – 428.6g = 571.4g
Q25. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
Ans.
Molar mass of Na2CO3
= (2 × 23) + 12 + (3 × 16)
= 106 g/mol
So, 1 mole of Na2CO3 means 106 g of Na2CO3
Thus, 0.5 mol of Na2CO3= (0.5 ×106)= 53 g of Na2CO3
Now, M denotes the molarity of the solution.
So,
0.5 M ofNa2CO3 = 0.5 mol/L of Na2CO3
Hence, 0.5 mol of Na2CO3 is in 1 L of water or 53 g ofNa2CO3 is in 1 L of water.
So, both are different in the way that the first one is just giving the mass of Na2CO3 whereas the second one is giving the mass dissolved per litre of solution.
Q26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Ans.
From a balanced chemical equation, we see that,
2 H2 + O2 → 2 H2O
So, 2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of water vapour.
Therefore, we can say that 10 volumes of dihydrogen will react with five volumes of dioxygen to produce 10 volumes of water vapour.
Q27. Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
Ans.
(i) 28.7 pm
We know that,
1 pm = 10−12m
Therefore, 28.7 pm = (28.7 ×10−12) m = 2.87 ×10−11 m
(ii) 15.15 pm
We know that,
1 pm = 10−12m
Therefore, 15.15 pm = (15.15 ×10−12) m = 1.515 ×10−11 m
(iii) 25365 mg
We know that,
1 mg= 10−6 kg
Therefore, 25365 mg = (25365×10−6) kg = 2.5365×10−2kg
Q28. Which one of the following will have the largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g ofCl2(g)
Ans.
(i) 1 g of Au (s)
Since, 197g of Au contains 6.022×1023 atoms of Au.
Therefore, 1g of Au contains (6.022×1023)/197 = 3.06×1021 atoms of Au.
(ii) 1 g of Na (s)
Since, 23g of Na contains 6.022×1023atoms of Na.
Therefore, 1g of Na contains (6.022×1023)/23 = 26.2×1021 atoms of Na.
(iii) 1 g of Li (s)
Since, 7g of Li contains 6.022×1023atoms of Li.
Therefore, 1g of Li contains (6.022×1023)/7 = 86 ×1021 atoms of Li
(iv)1 g ofCl2(g)
Since, 71g of Cl2contains 6.022×1023 atoms of Cl2.
Therefore, 1g of Cl2contains (6.022×1023)/71 = 8.48×1021molecules of Cl2 = 16.96×1021atoms of Cl
Therefore, 1 g of Li (s) will have the largest no. of atoms.
Q29. 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).
Ans.
We know the mole fraction of ethanol is defined as the ratio of the number of moles of ethanol to the total number of moles present in the solution.
Now,1 L of water contains 1000/18=55.55 moles of water.
Given, the mole fraction of ethanol is 0.040.
Let n be the number of moles of ethanol.
So we have, 0.040=n/ (n+55.55)
n=0.040n+2.222
n=(2.222)/(1−0.040)=2.31 moles.
As 1 L of solution contains 2.31 moles of ethanol, the molarity of a solution is 2.31 M.
Q30. What will be the mass of one 12C atom in g?
Ans.
We know that the molar mass of Carbon is 12g/mol
So, we can say that,
6.022 × 1023atoms of carbon has mass = 12g
So, 1 atom of carbon has mass = 12 / (6.022× 1023) g
= 1.993 × 10-23g
Q31. How many significant figures should be present in the answer of the following calculations?
(i) (0.02856× 298.15× 0.112) / 0.5785(ii) 5 × 5.364 (iii) 0.0125 + 0.7864 + 0.0215
Ans.
(i)(0.02856×298.15×0.112) / 0.5785
Here, the least precise number is = 0.112
Therefore, no. of significant numbers= no. of significant numbers in 0.112
= 3
(ii) 5 × 5.364
Here we have got the least precise number = 5.364
Therefore, no. of significant numbers= no. of significant numbers in 5.364
= 4
(iii) 0.0125 + 0.7864 + 0.0215
Here, the no. of significant numbers in the answer will be 4 as the least no. of decimal place in each term is 4.
Q32. Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:
| Isotope | Molar mass | Abundance |
| 36Ar | 35.96755g mol–1 | 0.337 % |
| 38Ar | 37.96272g mol–1 | 0.063 % |
| 40Ar | 39.9624g mol–1 | 99.600 % |
Ans.
Molar mass of Argon:
= [
{35.96755× (0.337/100)}
+
{37.96272× (0.063/100)}
+
{39.9624× (99.600/100)}]
= [0.121 + 0.024 + 39.802]g/mol
= 39.947 g/mol
Q33. Calculate the number of atoms in each of the following
(i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He
Ans.
(i)We know that,
1 mole of Ar =6.023×1023atoms of Ar
Therefore, 52 moles of Ar = 52 ×6.023×1023atoms of Ar
=3.131×1025atoms of Ar
(ii) 52 u of He
We know that,
4 u of He = 1 atom of He
1 u of He =1/4atom of He
Or, 52 u of He =52/4atoms of He= 13 atoms of He
(iii) 52 g of He
We know that,
4 g of He =6.023×1023atoms of He
52 g of He = (6.023×1023×52)/4atoms of He
=7.829×1024atoms of He
Q34. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Find:
(i) Empirical formula
(ii) Molar mass of the gas, and
(iii) Molecular formula
Ans.
Since, 3.38 g of carbon dioxide are obtained, the mass of C in the sample of welding fuel gas can be obtained as
(12/44)×3.38=0.92 g.
Also, 0.690 g of water are obtained, so, the mass of hydrogen present in the weldingsample is
(2/18)×0.690=0.077g.
The percentage of C is {0.92/(0.92+0.077)}×100=92.3 %.
The percentage of H is {0.077/(0.92+0.077)}×100=7.7%.
(i) So we get no. of moles of Carbon =92.3/12=7.7.
no. of moles of hydrogen =7.7/1=7.7.
Hence, the mole ratio C:H=7.7/7.7=1:1.
Therefore, we can say the empirical formula of the welding fuel gas is CH.
(ii) The weight of 10.0 L of gas at N.T.P is 11.6 g.
The weight of 22.4 L of gas at N.T. P = (11.6/10.0)×22.4=26g/mol
(iii) Empirical formula(CH) mass is 12+1=13.
Molecular mass is 26 (calculated above).
The ratio of the molecular mass to empirical formula mass is 26/13=2.
Therefore, we get molecular formula as 2(CH)=C2H2.
Q35. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) → CaCl2(aq) + CO2 (g) + H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
Ans.
The meaning of 0.75 M HCL is that 0.75mol of HCL is present in 1 L of solution.
Now, 1 mol of HCl = (1+35.5) g = 36.5g
∴0.75 mol of HCl = (36.5 ×0.75)g = 27.375g
So, 1000ml of HCL contains 27.375gofHCl.
Thus, 25ml of HCL contains (25 ×27.375)/1000 = 0.684g of HCl.
Now, from eq. we get that 2 mols of HCL react completely with 1 mol of CaCO3
Or, (2×36.5) g of HCL react with 100g of CaCO3(molar mass of CaCO3is100)
∴0.684g of HCL will react with (0.684×100) /(2×36.5) = 0.937g of CaCO3
Q36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction:
4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Ans.
In this question, firstly we will find out the mass of one mole of MnO2.
Mass of one mole of MnO2= 55+ 2 ×16 = 87g
Also, from the above eq., it is clear that 1 mole of MnO2reacts with 4 moles of HCL.
So, mass of 4 moles of HCL = 4× (1+35.5) = 146g
So, 87g of MnO2reacts with 146g of HCL.
Therefore, 5g of MnO2reacts with (5×146)/87 = 8.4g HCL.