Question

What is the concentration of a solution made by diluting 1.00 mL of concentrated HCl (12.1...

What is the concentration of a solution made by diluting 1.00 mL of concentrated HCl (12.1 M) to 100.0 mL?

A) 0.116 M
B) 0.242 M
C) 0.121 M
D) 0.0605 M
E) 0.100 M

Homework Answers

Answer #1

use dilution formula

M1*V1 = M2*V2

Here:

M1 is molarity of solution before dilution

M2 is molarity of solution after dilution

V1 is volume of solution before dilution

V2 is volume of solution after dilution

we have:

M1 = 12.1 M

V1 = 1.0 mL

V2 = 100.0 mL

we have below equation to be used:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (12.1*1)/100

M2 = 0.121 M

Answer: C

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
describe the steps to take in diluting a solution of 0.500M HCL to make 250 ml....
describe the steps to take in diluting a solution of 0.500M HCL to make 250 ml. of 0.100 M HCL.
if a solution is made by mixing 100.0 mL of 0.500 M NH3 with 100.0 mL...
if a solution is made by mixing 100.0 mL of 0.500 M NH3 with 100.0 mL of 0.100 M HCl, calculate its pH value (K b for NH3 = 1.8 x 10 –5 )
a) 30.0 ml of an HCl solution of unknown molarity was titrated with .200 M NaOH....
a) 30.0 ml of an HCl solution of unknown molarity was titrated with .200 M NaOH. Phenolphthalien was used as the indicator. The titrated solution turned a very pale pink after 21.8 mL of NaOH was added. What was the initial molarity of the HCl? b)Consider the following three titrations: 100.0 mL of 0.100 M CH3NH2 (Kb = 4.4x10-4) titrated with 0.100 M HCl 100.0 mL of 0.100 M NH3 (Kb = 1.8x10-5) titrated with 0.100 M HCl 100.0 mL...
1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that is...
1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that is 0.100 M HNO2 and .190 M NaNO2. How many moles of HNO2 and NaNO2 remain in solution after of the HCl
6. a. Calculate the pH of a solution prepared by diluting 2.1 mL of 2M HCl...
6. a. Calculate the pH of a solution prepared by diluting 2.1 mL of 2M HCl to a final volume of 100 mL with water. b. What is the pH of a solution that contains 0.5 M acetic acid and 0.25 M sodium acetate? The pKa for acetic acid is 4.7 Please explain why we do the steps in the calculations also.
a) What is the resulting concentration (M) of the solution containing 0.25 mL of 6M HCl...
a) What is the resulting concentration (M) of the solution containing 0.25 mL of 6M HCl and 12.50 mL distilled water? (0.0.75 pt.) b) What is the pH of this solution containing the HCl? (0.05 pt.) c) If 0.25 mL of 6M HCl is added in 4 mL of 0.2 M ClO- buffer solution, what is the final pH of the buffer (Ka HClO = 3.0 x 10-8)? (0.075 pt) d) If the Ka of the HClO is 3.0 x...
You drop 0.050 g of Mg chips into 100.0 mL of 1.00 M HCl and the...
You drop 0.050 g of Mg chips into 100.0 mL of 1.00 M HCl and the temperature of the solution increases from 22.21 °C to 24.46 °C. Assume that cs (solution) - 4.20 J/g °C and that the density of the solution is 1.00 g/mL. What is the enthalpy of the reaction per mole of Mg? Mg (s) + 2 HCl (aq) ----> H2 (g) + MgCl2 (aq)
The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl....
The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH after 7.6 mL of acid are added ______ Find the pH after 20 mL of acid are added _______
The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl....
The dibasic compound B (pKb1 = 4.00, pKb2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH after 3.1 mL of acid are added. Find the pH after 20 mL of acid are added.
calculate the concentration of free Ni2+ in a solution prepared by mixing 0.100 mol of ethylenediamine...
calculate the concentration of free Ni2+ in a solution prepared by mixing 0.100 mol of ethylenediamine plus 1.00 mL of 0.0100 M Ni2+ and diluting to 1.00 L with dilute base