Question

Answer the following questions in your subject DPQ notebook.

1. What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?
2. The empirical formula and molecular mass of a compound are CH2O and 180 g respectively. What will be the molecular formula of the compound ? (Given Molar mass: C = 12, H = 1, O = 16)
3. 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 (1) + MnCl2 (aq) + Cl2 (g)

How many grams of HCl react with 21.75 g of manganese dioxide ? (Given Molar mass, Mn = 55, O = 16, Cl = 35.5, H = 1) 4) Calculate the concentration of nitric acid in mole per litre in a sample which has a density,1.41 g/mL and the mass percent of nitric acid in it being 88.83%. (Molar mass H =1, N = 14, O = 16, Na = 23, Cl = 35.5) 5) A compound on analysis found to contain following percentage composition : Na = 43.4%, C = 11.4% and O = 45.3% Given: the relative molecular mass of the compound is 106, atomic mass Na=23, C=12, O=16 Answer the following questions: a. Define empirical formula? b. Calculate empirical and molecular formula of the compound according to the data given. 6) What are the prefixes and symbols of 10-9, 10-6, 10-12, 103, 10-15 and 1024? 7) Define molality and molarity of a solution. Out of molarity and molality which will you prefer to preserve a solution for a longer duration and wh ? 8) (a) 50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g). Calculate the amount of NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation. (b) What is a limiting reagent? 9) (a) What do you mean by significant figures? (b) Round up the following upto three significant figures: (i) 0.0048 (ii) 234,000 (iii) 34.459 (iv) 10.351

Answer 1

Here are the answers to the questions:

1) What will be the molarity of a solution, which contains 5.85 g of NaCl(s) per 500 mL?

• Molar mass of NaCl: $23.0 \, (\text{Na}) + 35.5 \, (\text{Cl}) = 58.5 \, \text{g/mol}$
• Moles of NaCl: $\frac{5.85 \, \text{g}}{58.5 \, \text{g/mol}} = 0.1 \, \text{mol}$
• Volume of solution in Liters: $500 \, \text{mL} = 0.5 \, \text{L}$
• Molarity (M): $\frac{\text{Moles of solute}}{\text{Volume of solution (L)}} = \frac{0.1 \, \text{mol}}{0.5 \, \text{L}} = 0.2 \, \text{M}$

2) The empirical formula and molecular mass of a compound are CH2O and 180 g respectively. What will be the molecular formula of the compound ? (Given Molar mass: C = 12, H = 1, O = 16)

• Empirical formula mass of CH2O: $1 \times 12 \, (\text{C}) + 2 \times 1 \, (\text{H}) + 1 \times 16 \, (\text{O}) = 12 + 2 + 16 = 30 \, \text{g/mol}$
• Determine 'n': $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{180 \, \text{g/mol}}{30 \, \text{g/mol}} = 6$
• Molecular formula: $n \times (\text{Empirical formula}) = 6 \times (\text{CH}_2\text{O}) = \text{C}_6\text{H}_{12}\text{O}_6$

3) Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction $4 \text{HCl (aq)} + \text{MnO}_2 \text{ (s)} \rightarrow 2\text{H}_2\text{O (l)} + \text{MnCl}_2 \text{ (aq)} + \text{Cl}_2 \text{ (g)}$ How many grams of HCl react with 21.75 g of manganese dioxide ? (Given Molar mass, Mn = 55, O = 16, Cl = 35.5, H = 1)

• Molar mass of MnO2: $55 \, (\text{Mn}) + 2 \times 16 \, (\text{O}) = 55 + 32 = 87 \, \text{g/mol}$
• Molar mass of HCl: $1 \, (\text{H}) + 35.5 \, (\text{Cl}) = 36.5 \, \text{g/mol}$
• Moles of MnO2: $\frac{21.75 \, \text{g}}{87 \, \text{g/mol}} = 0.25 \, \text{mol}$
• From the balanced equation: $1 \, \text{mol of MnO}_2$ reacts with $4 \, \text{mol of HCl}$.
• Moles of HCl required: $0.25 \, \text{mol MnO}_2 \times \left(\frac{4 \, \text{mol HCl}}{1 \, \text{mol MnO}_2}\right) = 1.0 \, \text{mol HCl}$
• Mass of HCl: $1.0 \, \text{mol} \times 36.5 \, \text{g/mol} = 36.5 \, \text{g}$

4) Calculate the concentration of nitric acid in mole per litre in a sample which has a density,1.41 g/mL and the mass percent of nitric acid in it being 88.83%. (Molar mass H =1, N = 14, O = 16, Na = 23, Cl = 35.5)

• Molar mass of HNO3: $1 \, (\text{H}) + 14 \, (\text{N}) + 3 \times 16 \, (\text{O}) = 1 + 14 + 48 = 63 \, \text{g/mol}$
• Assume 100 g of solution:
  • Mass of HNO3 = $88.83 \, \text{g}$
  • Moles of HNO3 = $\frac{88.83 \, \text{g}}{63 \, \text{g/mol}} \approx 1.410 \, \text{mol}$
• Volume of 100 g solution:
  • Volume = $\frac{\text{Mass}}{\text{Density}} = \frac{100 \, \text{g}}{1.41 \, \text{g/mL}} \approx 70.92 \, \text{mL}$
  • Volume in Liters = $\frac{70.92 \, \text{mL}}{1000 \, \text{mL/L}} = 0.07092 \, \text{L}$
• Molarity (M): $\frac{1.410 \, \text{mol}}{0.07092 \, \text{L}} \approx 19.88 \, \text{M}$

5) A compound on analysis found to contain following percentage composition : Na = 43.4%, C = 11.4% and O = 45.3% Given: the relative molecular mass of the compound is 106, atomic mass Na=23, C=12, O=16 Answer the following questions:

a. Define empirical formula?

• The empirical formula of a compound is the simplest whole number ratio of atoms present in a compound.

b. Calculate empirical and molecular formula of the compound according to the data given.

Element	Percentage (%)	Atomic Mass (g/mol)	Moles (Percentage/Atomic Mass)	Mole Ratio (Divide by smallest)	Simplest Whole Number Ratio
Na	43.4	23	$43.4 / 23 = 1.887$	$1.887 / 0.95 = 1.986 \approx 2$	2
C	11.4	12	$11.4 / 12 = 0.95$	$0.95 / 0.95 = 1$	1
O	45.3	16	$45.3 / 16 = 2.831$	$2.831 / 0.95 = 2.98 \approx 3$	3

• Empirical Formula: $\text{Na}_2\text{CO}_3$
• Empirical Formula Mass: $2 \times 23 \, (\text{Na}) + 1 \times 12 \, (\text{C}) + 3 \times 16 \, (\text{O}) = 46 + 12 + 48 = 106 \, \text{g/mol}$
• Determine 'n': $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{106 \, \text{g/mol}}{106 \, \text{g/mol}} = 1$
• Molecular Formula: $n \times (\text{Empirical formula}) = 1 \times (\text{Na}_2\text{CO}_3) = \text{Na}_2\text{CO}_3$

6) What are the prefixes and symbols of 10-9, 10-6, 10-12, 103, 10-15 and 1024?

• $10^{-9}$: nano (n)
• $10^{-6}$: micro (µ)
• $10^{-12}$: pico (p)
• $10^{3}$: kilo (k)
• $10^{-15}$: femto (f)
• $10^{24}$: yotta (Y)

7) Define molality and molarity of a solution. Out of molarity and molality which will you prefer to preserve a solution for a longer duration and wh ?

• Molality (m): It is defined as the number of moles of solute dissolved per kilogram (kg) of solvent. $\text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}$
• Molarity (M): It is defined as the number of moles of solute dissolved per litre (L) of solution. $\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$
• Preference for preservation: Molality is preferred to preserve a solution for a longer duration. This is because molality is independent of temperature, as it is based on the masses of solute and solvent, which do not change with temperature. Molarity, on the other hand, depends on the volume of the solution, which changes with temperature (due to thermal expansion or contraction), thus altering its concentration.

8) (a) 50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g). Calculate the amount of NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this situation. (b) What is a limiting reagent?

a. Calculate the amount of NH3 (g) formed. Identify the limiting reagent.

• Balanced chemical equation: $\text{N}_2 \text{ (g)} + 3\text{H}_2 \text{ (g)} \rightarrow 2\text{NH}_3 \text{ (g)}$
• Molar masses: $\text{N}_2 = 28 \, \text{g/mol}$, $\text{H}_2 = 2 \, \text{g/mol}$, $\text{NH}_3 = 17 \, \text{g/mol}$
• Given masses: $\text{N}_2 = 50.0 \, \text{kg} = 50,000 \, \text{g}$, $\text{H}_2 = 10.0 \, \text{kg} = 10,000 \, \text{g}$
• Moles of reactants:
  • Moles of $\text{N}_2 = \frac{50,000 \, \text{g}}{28 \, \text{g/mol}} \approx 1785.71 \, \text{mol}$
  • Moles of $\text{H}_2 = \frac{10,000 \, \text{g}}{2 \, \text{g/mol}} = 5000 \, \text{mol}$
• Identify the limiting reagent:
  • According to the stoichiometry, $1 \, \text{mol of N}_2$ reacts with $3 \, \text{mol of H}_2$.
  • To react completely with $1785.71 \, \text{mol of N}_2$, moles of $\text{H}_2$ needed = $1785.71 \times 3 = 5357.13 \, \text{mol}$.
  • Since only $5000 \, \text{mol of H}_2$ is available, which is less than the required amount, $\text{H}_2$ is the limiting reagent.
• Calculate amount of NH3 formed (based on limiting reagent H2):
  • From the equation, $3 \, \text{mol of H}_2$ produces $2 \, \text{mol of NH}_3$.
  • Moles of $\text{NH}_3$ produced = $5000 \, \text{mol H}_2 \times \left(\frac{2 \, \text{mol NH}_3}{3 \, \text{mol H}_2}\right) \approx 3333.33 \, \text{mol NH}_3$
  • Mass of $\text{NH}_3$ produced = $3333.33 \, \text{mol} \times 17 \, \text{g/mol} \approx 56666.61 \, \text{g} = 56.67 \, \text{kg}$

b. What is a limiting reagent?

• A limiting reagent (or limiting reactant) is the reactant in a chemical reaction that is completely consumed first, thereby stopping the reaction and limiting the maximum amount of product that can be formed.

9) (a) What do you mean by significant figures? (b) Round up the following upto three significant figures: (i) 0.0048 (ii) 234,000 (iii) 34.459 (iv) 10.351

a. What do you mean by significant figures?

• Significant figures are the digits in a measured or calculated quantity that are considered to be reliable and contribute to the precision of the measurement. They include all non-zero digits, zeros between non-zero digits, and trailing zeros when a decimal point is present. Leading zeros (zeros before non-zero digits) are not significant.

b. Round up the following upto three significant figures:

• (i) 0.0048: The significant figures are 4 and 8. To express it with three significant figures, we add a trailing zero.
  • Rounded value: 0.00480
• (ii) 234,000: The significant figures are 2, 3, and 4. The trailing zeros are not significant without a decimal point. Thus, it already has three significant figures.
  • Rounded value: 234,000
• (iii) 34.459: The first three significant figures are 3, 4, and 4. The next digit (5) requires rounding up the last significant digit.
  • Rounded value: 34.5
• (iv) 10.351: The first three significant figures are 1, 0, and 3. The next digit (5) requires rounding up the last significant digit.
  • Rounded value: 10.4

Answer 2

Here are the answers to the questions:

1. Molarity: Calculated moles of NaCl from mass and molar mass, then divided by the volume of solution in liters.
2. Molecular Formula: Determined 'n' by dividing the given molecular mass by the calculated empirical formula mass. Multiplied the empirical formula by 'n' to get the molecular formula.
3. Stoichiometry: Converted the mass of MnO2 to moles. Used the mole ratio from the balanced equation to find moles of HCl required. Converted moles of HCl to grams.
4. Concentration (Molarity from % and Density): Assumed 100g of solution to find the mass of HNO3. Calculated moles of HNO3. Used density to find the volume of 100g solution. Divided moles by volume (in L) to get molarity.
5. Empirical and Molecular Formula from % Composition:
   • a. Defined empirical formula as the simplest whole-number ratio of atoms.
   • b. Converted percentage composition to moles for each element. Divided by the smallest number of moles to find the simplest mole ratio, leading to the empirical formula. Calculated empirical formula mass. Used the given molecular mass to find 'n' and then derived the molecular formula.
6. Prefixes and Symbols: Listed the standard SI prefixes and symbols corresponding to the given powers of 10.
7. Molality and Molarity Definitions and Preference: Defined both terms. Explained that molality is preferred for long-term preservation because it is temperature-independent (based on mass), unlike molarity (based on volume).
8. Limiting Reagent and Product Yield:
   • a. Converted masses of reactants to moles. Used stoichiometric ratios to determine which reactant was limiting (H2 in this case). Calculated the moles of product (NH3) formed based on the limiting reagent, then converted to mass.
   • b. Defined limiting reagent as the reactant completely consumed first, limiting product formation.
9. Significant Figures:
   • a. Defined significant figures as reliable digits indicating precision, including specific rules for non-zero digits and zeros.
   • b. Applied standard rounding rules to round the given numbers to three significant figures.