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A 251 mL sample of 0.45 M HCl is added to 455 mL of distilled water. What is the molarity of the final...

10) A 251 mL sample of 0.45 M HCl is added to 455 mL of distilled water. What is the molarity of the final solution? (Hint: Find the number of moles of HCl and the total volume of the final solution)
11) How many millilitres of 0.250 M KMnO4 are needed to deliver 0.00450 moles of KMnO4 in a titration?
12) How many fluorine atoms are in 750.0 mL of a 0.500M HF solution?
13) How many moles of NH3 are in a 3.0 L solution of 0.23M NH3?
14) Suppose you needed to prepare a 100.0 mL of 1.05 M NaOH using 1.50 M NaOH, distilled water, and a 100 mL graduated cylinder. How would you do this?
15) Find the mole fraction of glucose, C6H12O6, in a solution that contains 2.1 moles of glucose and 55.49 moles of water.
16) A rigid 5.5 L sealed vessel contains 0.350 moles N2(g), 0.125 moles Ar(g), and 0.110 moles He(g). Find the mole fraction of each gas.
17) A gaseous solution contains 41.0% O2 and 59.0% N2 by mass. Find the mole fraction of each substance in the solution.

Answer

  1. Calculate the moles of HCl:
    Moles = 0.45 M × 0.251 L = 0.11305 moles.
    Total volume = 251 mL + 455 mL = 706 mL = 0.706 L.
    Molarity = 0.11305 moles / 0.706 L = 0.160 M.

  2. Moles of KMnO4 = 0.00450 moles.
    Volume = Moles / Molarity = 0.00450 moles / 0.250 M = 0.018 L = 18 mL.

  3. Moles of HF = 0.500 M × 0.750 L = 0.375 moles.
    Fluorine atoms = 0.375 moles × 6.022 × 1023 atoms/mole = 2.26 × 1023 atoms.

  4. Moles of NH3 = 0.23 M × 3.0 L = 0.69 moles.

  5. Use dilution formula: M1V1 = M2V2.
    1.50 M × V1 = 1.05 M × 0.100 L.
    V1 = (1.05 × 0.100) / 1.50 = 0.070 L = 70 mL.
    Mix 70 mL of 1.50 M NaOH with 30 mL of distilled water.

  6. Total moles = 2.1 + 55.49 = 57.59 moles.
    Mole fraction of glucose = 2.1 / 57.59 = 0.0365.

  7. Total moles = 0.350 + 0.125 + 0.110 = 0.585 moles.
    Mole fraction of N2 = 0.350 / 0.585 = 0.598.
    Mole fraction of Ar = 0.125 / 0.585 = 0.214.
    Mole fraction of He = 0.110 / 0.585 = 0.188.

  8. Assume 100 g of solution: 41 g O2 and 59 g N2.
    Moles of O2 = 41 g / 32 g/mol = 1.281 moles.
    Moles of N2 = 59 g / 28 g/mol = 2.107 moles.
    Total moles = 1.281 + 2.107 = 3.388 moles.
    Mole fraction of O2 = 1.281 / 3.388 = 0.378.
    Mole fraction of N2 = 2.107 / 3.388 = 0.622.

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