1.) PLEASE ANSWER ALL THE FOLLOWING A.) Complete the row for KNO3. Enter your an

Question

1.) PLEASE ANSWER ALL THE FOLLOWING

A.) Complete the row for KNO3.Enter your answers numerically separated by a comma.

B.) Complete the row for NaHCO3.Enter your answers numerically separated by a comma.

C.)Complete the row for C12H22O11.Enter your answers numerically separated by a comma.

2.) 198 mL sample of a 1.3 M sucrose solution is diluted to 300 mL .What is the molarity of the diluted solution? Express your answer using two significant figures.

3.) Determine the volume of 0.165 M NaOH solution required to neutralize each of the following samples of hydrochloric acid. The neutralization reaction is:
NaOH(aq)+HCl(aq)H2O(l)+NaCl(aq)

a.) 30 mL of a 0.165 M HCl solution. Express your answer using two significant figures.

b.) 75 mL of a 0.045 M HCl solution .Express your answer using two significant figures.

c.) 185 mL of a 0.855 M HCl solution . Express your answer using three significant figures.

4.)A 11.0 mL sample of an unknown H3PO4solution requires 118 mL of 0.130 M KOH to completely react with the H3PO4 according the following reaction. What was the concentration of the unknown H3PO4 solution? H3PO4(aq)+3KOH(aq)3H2O(l)+K3PO4(aq)

5.)Compute the molality of each of the following

a. 0.55 mol solute; 0.350 kg solvent. Express your answer using two significant figures.

b. 0.922 mol solute; 0.250 kg solvent .Express your answer using three significant figures.

c. 0.010 mol solute; 25.1 g solvent .Express your answer using two significant figures.

MASS MOL VOLUME SOLUTE SOLUTE SOLUTE SOLUTION MOLARITY KNO3 20.5 g ? 129.0 ml ? NaHCO3 ? ? 240.0 ml 0.140 M C12H22O11 60.38g ? ? 0.150M

Explanation / Answer

A) Moles of KNO3 = mass KNO3/mol wt KNO3 = 20.5/101 = 0.2029 moles

Molarity = moles/volume = 0.2029 moles/0.129 L = 1.57M

B) moles of NaHCO3 = Molarity * Volume = 0.140 * .240 = 0.0336 moles

Moles of NaHCO3 = mass NaHCO3/mol wt of NaHCO3

Mass of NaHCO3 = 0.0336 moles * 84 g/mol = 2.822 g

C) Moles of C12H22O11 = mass/mol wt = 60.38/342 = 0.1765 moles

Molarity = moles/volume

Volume = moles/molarity = 0.1765/0.150 = 1.18 L

2) Use equation:

M1V1 = M2V2

M2 = M1V1/V2 = 198*1.3/300 = 0.86M

3) As per the neutralization reaction: NaOH and HCl react in the ratio 1:1

# moles of HCl = # moles of NaOH

a) # moles of HCl in 30ml of 0.165 M = 0.030 * 0.165 = 0.00495 moles = # moles of NaOH

Vol of NaOH = moles / molarity = 0.00495 / 0.165 = 0.030L

b) # moles of HCl= 0.075 * 0.045 = 0.00338 moles

Vol naoH = 0.00338 / 0.165 = 0.021L

c) moles of HCl = 0.185 * 0.855 = 0.1582 moles

vol of NaOH = 0.1582/0,165= 0.96 M

4) # moles of KOH in 118 ml of 0.130 M = .118 * 0.130 = 0.01534 moles

1 moles of H3PO4 reacts with 3 moles of KOH

Therefore # moles of H3PO4 = 1/3 moles of KOH = 0.01534/3 = 0.00511 moles

i.e 0.00511 moles of H3PO4 is present in 11 ml = 0.011L of the solution

Concentration = molarity = moles of H3PO4/ lit = 0.00511/0.011 = 0.46M

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