Question

Calculate the number of moles of solute in 83.93 mL of 0.1735 M K2Cr2O7(aq).

Calculate the number of moles of solute in 83.93 mL of 0.1735 M K2Cr2O7(aq).

Homework Answers

Answer #1

We know that the formula for the molarity is

Molar concentration or molarity is most commonly expressed in units of moles of solute per litre of solution. For use in broader applications, it is defined as amount of solute per unit volume of solution, or per unit volume available to the species, represented by lowercase c:

Here, n is the amount of the solute in moles, N is the number of molecules present in the volume V (in litres), the ratio N/V is the number concentration C, and NA is theAvogadro constant, approximately 6.022×1023 mol−1.
Or more simply: 1 molar = 1 M = 1 mole/litre.

So the number of moles of solute = Molarity * Volume in Litre

As given in the question Volume = 83.93 ml = 0.08393 L

and Molarity = 0.1735 M

So  moles of solute of K2Cr2O7 = 0.08393 * 0.1735 = 0.01456 moles

Thank you

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
What mass in grams of solute is needed to prepare a)0.250 L of 0.167 M K2Cr2O7...
What mass in grams of solute is needed to prepare a)0.250 L of 0.167 M K2Cr2O7 and b) 625mL of 0.0200 M KMnO4?
Calculate the moles of IO3- dispensed if you have 21.56 mL KIO3 M 1.0 and IO3-...
Calculate the moles of IO3- dispensed if you have 21.56 mL KIO3 M 1.0 and IO3- (aq) +8I- (aq) +H+ (aq) -> 3I3- (aq) + 3H2O(l). Then calculate the moles of I3- produced. Then calculate the moles of C6H8O6 in the sample. C6H8O6 (aq) + I3- (aq) + H2O(l) -> C6H8O7(aq) + 3I- (aq) + 2H+ (aq) Thank you!
What mass in grams of solute is needed to prepare 575 mL of 1.90×10−2 M KMnO4?...
What mass in grams of solute is needed to prepare 575 mL of 1.90×10−2 M KMnO4? What mass in grams of solute is needed to prepare 0.390 L of 0.229 M K2Cr2O7?
How many moles of solute are present in the following solutions: 1.) 145 mL of 0.32...
How many moles of solute are present in the following solutions: 1.) 145 mL of 0.32 M NaNO3 2.) 430 mL of 1.3 M HNO3
Part 1. using the ZN sample mass (1.948), calculate the number of moles of ZN (m)...
Part 1. using the ZN sample mass (1.948), calculate the number of moles of ZN (m) used in the reaction. NOTE: in the experiment it said to dispense approximately 2 grams of ZN and 10 ml HCl Part 2. Using the number of moles of Zn, and the mole ratio in the reaction, Zn (m) + 2 HCl (aq) - ZnCl2 (aq) + H2 (g) determine the number of h2 produced Part 3. The pressure observed after reaction with Zn...
35. How many moles of solute particles are present in 100.0 mL of 2.50 M (NH4)3PO4?...
35. How many moles of solute particles are present in 100.0 mL of 2.50 M (NH4)3PO4? A) 0.100 mol B) 0.250 mol C) 0.500 mol D) 0.750 mol E) l.00 mol Ans: E.. how do you get the answer?
A 430. mL solution of 0.230 M HBr(aq) is mixed with a solution of 225 mL...
A 430. mL solution of 0.230 M HBr(aq) is mixed with a solution of 225 mL of 0.350 M HClO4(aq). How many moles of H+(aq) are present in the final solution?
A 15.00 mL of 0.250 M HNO3 (aq) is titrated with 30.0 mL of 0.175 M...
A 15.00 mL of 0.250 M HNO3 (aq) is titrated with 30.0 mL of 0.175 M KOH (aq). d.) After neutralization, how many moles of excess reactant remain? e.) What is the pH for this titration mixture?
a) Calculate the change in pH when 8.00 mL of 0.100 M HCl(aq) is added to...
a) Calculate the change in pH when 8.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq). b) Calculate the change in pH when 8.00 mL of 0.100 M NaOH(aq) is added to the original buffer solution.
Calculate the pH of the resulting solution if 17.0 mL of 0.170 M HCl(aq) is added...
Calculate the pH of the resulting solution if 17.0 mL of 0.170 M HCl(aq) is added to a)22.0 mL of 0.170 M NaOH(aq). (b) 27.0 mL of 0.220 M NaOH(aq).