Suppose you had 58.44 g of NaCl and you dissolved it in exactly 2.00 L of water. What would be the molality? 2. Calculate the molarity of 25.0 grams of KBr dissolved to 750.0 mL. 3. Eighty grams of glucose (C6H12O6, MW = 180. g/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity? 4. When 2.00 grams of KMnO4 (MW = 158.0 g/mol) is dissolved into 100.0 mL of solution, what normality results? How many grams of KMnO4 are needed to make 500.0 mL of a 0.200 N solution? 5. Ten grams of acetic acid (CH3COOH) is dissolved in 500.0 mL of solution. What is the molarity? 6. How many mL of solution will result when 15.0 g of H2SO4 is dissolved to make a 0.200 M solution? 7. Convert 0.169 Molar HC2H3O2 to % (m/v). 8. One gram of sodium hydroxide pellets is dissolved in two hundred-fifty milliliters water. What is the molarity, molality and normality of the solution? Assume that the density of solution to be approximately that of the water and the density of sodium hydroxide pellets to be 2.13 g/cm³.
The Correct Answer and Explanation is :
Solutions and Explanations:
- Molality of NaCl solution: Given:
- Mass of NaCl = 58.44 g
- Volume of water = 2.00 L ≈ 2000 g (assume density of water = 1 g/mL) Steps:
- Moles of NaCl = ( \frac{58.44 \, \text{g}}{58.44 \, \text{g/mol}} = 1.00 \, \text{mol} )
- Molality ((m)) = ( \frac{\text{moles of solute}}{\text{mass of solvent in kg}} )
- ( m = \frac{1.00}{2.00} = 0.500 \, \text{mol/kg} ) Answer: ( 0.500 \, m )
- Molarity of KBr solution: Given:
- Mass of KBr = 25.0 g
- Volume of solution = 750.0 mL = 0.750 L
- Molar mass of KBr = 119.0 g/mol Steps:
- Moles of KBr = ( \frac{25.0}{119.0} = 0.210 \, \text{mol} )
- Molarity ((M)) = ( \frac{\text{moles of solute}}{\text{volume of solution in L}} )
- ( M = \frac{0.210}{0.750} = 0.280 \, \text{M} ) Answer: ( 0.280 \, M )
- Molarity of glucose solution: Given:
- Mass of glucose = 80.0 g
- Volume of solution = 1.00 L
- Molar mass of glucose = 180. g/mol Steps:
- Moles of glucose = ( \frac{80.0}{180.} = 0.444 \, \text{mol} )
- Molarity ((M)) = ( \frac{0.444}{1.00} = 0.444 \, \text{M} ) Answer: ( 0.444 \, M )
- Normality of KMnO(_4) solution: Given:
- Mass of KMnO(_4) = 2.00 g
- Volume = 100.0 mL = 0.100 L
- Molar mass = 158.0 g/mol
- Equivalent factor = 5 (KMnO(_4) in acidic medium) Steps:
- Moles of KMnO(_4) = ( \frac{2.00}{158.0} = 0.0127 \, \text{mol} )
- Normality ((N)) = ( \text{Molarity} \times \text{Equivalent factor} )
- ( N = \frac{0.0127}{0.100} \times 5 = 0.635 \, N ) Grams needed for 0.200 N solution:
- ( 0.200 = \frac{\text{grams of KMnO}_4 \times 5}{158.0 \times 0.500} )
- Grams = ( 3.16 \, \text{g} ) Answer: ( N = 0.635, \, \text{Grams needed} = 3.16 \, \text{g} )
- Molarity of acetic acid: Given:
- Mass of CH(_3)COOH = 10.0 g
- Volume = 500.0 mL = 0.500 L
- Molar mass = 60.0 g/mol Steps:
- Moles = ( \frac{10.0}{60.0} = 0.167 \, \text{mol} )
- ( M = \frac{0.167}{0.500} = 0.334 \, \text{M} ) Answer: ( 0.334 \, M )
- Volume for H(_2)SO(_4) solution: Given:
- Mass of H(_2)SO(_4) = 15.0 g
- Molar mass = 98.0 g/mol
- ( M = 0.200 \, \text{M} ) Steps:
- Moles = ( \frac{15.0}{98.0} = 0.153 \, \text{mol} )
- Volume = ( \frac{\text{moles}}{M} = \frac{0.153}{0.200} = 0.765 \, \text{L} ) Answer: ( 765 \, \text{mL} )
- Convert 0.169 M HC(_2)H(_3)O(_2) to % (m/v): Given:
- ( M = 0.169 \, \text{M} )
- Molar mass = 60.0 g/mol Steps:
- ( 0.169 \, \text{mol/L} \times 60.0 \, \text{g/mol} = 10.14 \, \text{g/L} )
- % (m/v) = ( \frac{10.14 \, \text{g}}{1000 \, \text{mL}} \times 100 = 1.01\% ) Answer: ( 1.01\% \, \text{(m/v)} )
- Sodium hydroxide solution: Given:
- Mass of NaOH = 1.00 g
- Volume = 250.0 mL = 0.250 L
- Molar mass = 40.0 g/mol Steps:
- Moles = ( \frac{1.00}{40.0} = 0.025 \, \text{mol} )
- Molarity = ( \frac{0.025}{0.250} = 0.100 \, \text{M} )
- Assume ( \text{mass of solution} = 250 \, \text{g} ) (water density)
- Molality = ( \frac{0.025}{0.250} = 0.100 \, \text{mol/kg} )
- Normality ((N)) = Molarity (1 equivalent) Answer: ( 0.100 \, M, \, 0.100 \, m, \, 0.100 \, N )
Explanation:
- The calculations follow the basic definitions of molarity, molality, and normality, with the molar mass and volume conversions ensuring accurate results.
- Assumptions such as water density simplify calculations without significantly affecting accuracy for dilute solutions.