Some Basic Concepts of Chemistry Test - 12

Precision refers to the closeness of various measurements for the same quantity and accuracy is the agreement of a particular value to the true value of the result. Results of student A are very close to 3g.

\(^0F\) = \(9\over5\)(\(^0C\)) + 32

\(^0C\) = \(5\over9\)(\(^0F\) - 32) = \(5\over9\)(200 - 32) = 93.3°C

No. of moles of NaCl = \(5.85\over 58.5\) = 0.1 mol/L

500ml = \(500\over 1000\) = 0.5 L

Molarity(M) = \(No.\,of\,moles\,of\,solute\over Vol.\,of\,solution\,in\,litres\)

Molarity = \(0.1\over 0.5\) = 0.2M

\(M_1V_1\) = \(M_2V_2\)

\(M_2={M_1V_1\over V_2}\) = \(500\times 5\over 1500\) = 1.66M

Number of atoms = Moles x \(N_A\)

Number of moles = \(wt.\over Mol.wt.\)

(a) 4 g He = \(4\over 4\) = 1 mole

(b) 46 g Na = \(46\over 23\) = 2 moles

(c) 0.40 g Ca = \(0.40\over 40\) = 0.01 mole

(d) 12 g He = \(12\over 4\) = 3 moles

Hence, 12 g of He contains the greatest number of atoms.

Concentration of glucose in blood = 0.9g/L

Molar mass of glucose = 180 g \(mol ^{-1}\)

0.9g = \(0.9\over180\) = \(1\over200\) moles

Molarity of glucose in blood = 0.005 M

No. of moles of HCL = \(18.25\over36.5\) = 0.5

Molality(m) = \(No.\,of\,moles\,of\,solute\over Mass\,of\,solvent\, in \,kg\)

Molality = \(0.5\times 1000\over 500\) = 1 m

Molarity = 0.02 M, Volume of sol.= 100 mL= 0.1 L

No. of moles of \(H_2SO_4\) = molarity x volume in litres

= 0.02 x 0.1 = 2 x \(10^{-3}\)

No. of molecules of \(H_2SO_4\) = 2 x \(10^{-3}\) x 6.022 x \(10^{23}\)

= 12.044 x \(10^{20}\) molecules

Mass percent of C in \(CO_2\) = \(Molar\,mass\, of\, carbon \times 100\over Molar\, mass \, of\, CO_2\)

\(\therefore\) % of C in \(CO_2\) = \(12\over 44\) x 100 = 27.27%

Molecular formula = n x (empirical formula)

n = \(Molecular\, mass\over Empirical\, formula\, mass\)

Empirical formula of ethene = \(CH_2\)

Empirical formula mass of ethene = 14 amu

= \(1\over2\) x molecular mass of ethene

Empirical formula shows that ethene has (C : H):: 1:2

1 mol of oxygen gas at STP = 6.022 x \(10^{23}\) molecules of oxygen.

1 mole \(O_2\) = 32 g/mol of \(O_2\)

Empirical formula = \(CH_2O\)

Empirical formula mass = 12+1+1+16 = 30g

Molecular mass = 180 g

n = \(Molar\,mass\over Empirical\,formula\, mass\) = \(180\over 30\) = 6

\(\therefore\) n = 6

So, molecular formula = n x \(CH_2O\) = 6 x \(CH_2O\) = \(C_6H_{12}O_6\)

Mass = Density x Volume

= 3.12 g m\(L^{–1}\) x 1.5 mL = 4.68 g = 4.7 g

The compound does not retain the physical properties of its constituent elements.

Carbon-12 is one of the isotopes of carbon and can be represented as \(^{12}C\). In this system, \(^{12}C\) is assigned a mass of exactly 12 atomic mass unit (amu) and mass of all other atoms are given relative to this standard. One atomic mass unit is defined as a mass exactly equal to one twelfth the mass of one carbon-12 atom.

\(H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O\)

0.1 M \(H_2SO_4\) = 0.1 mole of \(H_2SO_4\)

0.1 mole \(H_2SO_4\) reacts with 2 moles of NaOH

0.1 mole of NaOH will react with = \(0.1\over2\) moles of \(H_2SO_4\)

= 0.05 moles of \(H_2SO_4\)

Here NaOH is the limiting reactant

2 moles of NaOH produce 1 mole of \(Na_2SO_4\)

0.1 mole of NaOH = \(0.1\over2\) = 0.05 mole

mass of \(Na_2SO_4\) = 0.05 x (Molar mass of \(Na_2SO_4\))

= 0.05 x (46 + 32 + 64)

= 0.05 x 142 = 7.10 g

Volume of solution after mixing = 2L

Molarity of \(Na_2SO_4\) = \(0.05\over2\) = 0.025 mol \(L^{–1}\)

Zero at the end or right of a number are significant provided they are on the right side of the decimal point. For example, 0.200 g has three significant figures.

One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12 C isotope.

12g C = 1 mole of C and, and 23g Na = 1 mole of Na

28g \(N_2\) = 1 mole of nitrogen = 2 x 6.022 x \(10^{23}\) atoms of nitrogen

32g \(O_2\) = 1 mole of oxygen = 2 x 6.022 x \(10^{23}\) atoms of oxygen

(i) 20 g NaOH = \(20\over 40\) = 0.5 mol NaOH in 200 ml solution.

(ii) Molar concentration of NaOH = \(20\over 40\) = \(0.500\,mol\over 0.200\,L\) = 2.5 M

Molar concentration of KCl = \(0.5\,mol\over 0.200\,L\) = 2.5 M

According to the law of conservation, the mass of the reactants in a chemical reaction is equal to the mass of the product.

No. of atoms in the reactant side is not equal to the no. of atoms in the product side.

Number of molecules of oxygen in 16g of oxygen = \(16\over 32\) x 6.023 x \(10^{23}\) 

= 0.5 x 6.023 x \(10^{23}\)

0.5 moles is present in 14g of nitrogen and in 1.0 g \(H_2\). Hence they will also have 0.5 x 6.023  x \(10^{23}\)

iii) Number of molecules of \(N_2\) = \(14\over28\) x 6.023 x \(10^{23}\) = 0.5 x 6.023  x \(10^{23}\)

iv) Number of molecules of \(H_2\) = \(1\over2\) x 6.023 x \(10^{23}\) = 0.5 x 6.023  x \(10^{23}\)

According to this law, 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 the ratio of small whole numbers.

Mass percent = \(Mass\,of\,solute\over Mass\,of\,solution\) x 100

Mole Fraction: It is the ratio of number of moles of a particular component to the total number of moles of the solution.

\(CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)\)

16g of \(CH_4\) on complete combustion will give 36g of water.

Law of conservation of mass states that matter can neither be created nor destroyed. Avogadro proposed that equal volumes of gases at the same temperature and pressure should contain equal number of molecules.