Chemistry

posted by .

WHAT IS THE BOILING POINT OF THIS SOLUTION?

1.00x10^2g C10,H8,O6,S2 (1,5-naphthalenedisulfonic acid) in 1.00x10^2g H2O (nonionizing solute)

  • Chemistry -

    -6.25

  • Chemistry -

    -6.25

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. chemistry

    Compute the freezing point of this solution: 25.5 g C7H11NO7S (4-nitro-2-toluenesulfoinoic acid dihydrate) in 100 g H2O. (nonionizing solute)
  2. chemistry

    Compute the boiling point of this Solution: 25.5g C7H11NO7S (4-nitro-2-toluenesulfonoic acid dihydrate) in 1.00*10^2g H2O (nonionizing solute) The freezing point of H2O is lowered 1.86 Celsius per mole of solute. The boiling pint os …
  3. chemistry

    Compute the freezing point of this Solution: 25.5g C7H11NO7S (4-nitro-2-toluenesulfonoic acid dihydrate) in 1.00*10^2g H2O (nonionizing solute) The freezing point of H2O is lowered 1.86 Celsius per mole of solute. The boiling point …
  4. Chemistry

    Compute the boiling point of this solution: 1.00*10^2g C10H8O6S2 (1,5-naphthalenedisulfonic acid) in 1.00*10^2g H2O (nonionizing solute) The freezing point of H2O is Lowered 1.86 Celsius per mole of solute. The boiling point of H2O …
  5. Chemistry

    WHAT IS THE FREEZING POINT OF THIS SOLUTION?
  6. Chemistry

    WHAT IS THE BOILING POINT OF THIS SOLUTION?
  7. Gen Chemistry

    The following absorption data of a solution was collected in a cell with 1 cm path length. The absorption of the same substance in a solution with unknown concentration was 0.197. What is the concentration of the unknown substance?
  8. Chemistry

    A solution of 1.50g of solute dissolved in 25.0 mL of H2O at 25C has a boiling point of 100.95C. What is the molar mass of the solute if it is a nonvolatile non electrolyte and the solution behaves ideally. (d of H2O at 25C=0.997g/mL)?
  9. Chemistry

    what is the solubility in mol/L of insoluble Ca3(PO4)2 (Ksp=1.00x10^-26)in a solution containing .00100 M Na3PO4?
  10. Chemistry

    30.0 mL of 0.0200 mol/L NaOH are added to 70.0 mL of 0.0100 mol/L HCI. Calculate the [H3O+] and [OH─] of the resulting solution My answer: [OH-] = 1.00x10^-10 M; [H3O+] = 1.00x10^-4 M Teacher's answer: [OH-] = 1.00x10^-11 M; …

More Similar Questions