Question

1) a. What is the molar concentration (i.e., molarity) of sucrose (C12H22O11) if 150.0 g is dissolved in 250.0 mL of solution?

b.What is the molar concentration of methanol (CH3OH) if 125.0 mL is dissolved in enough water to make 15.0 L of solution? The density of methanol is 0.792 g/mL.

c.How many grams of NaOH are contained in 250.0 mL of a 0.400 M sodium hydroxide solution?

d.The drinking water standard for lead is 15 ppb (parts per billion). What is this in M? (See Appendix C – Solution Preparation)

e.What volume of 0.650 M sucrose must be diluted with water to prepare 250.0 mL of 0.423 M sucrose? (See Appendix C – Solution Preparation)

f.Which of the following solutions contains the largest concentration of chloride ions: Which contains the largest number of chloride ions? Show work. a. 100 mL of 0.30 M AlCl3 b. 50.0 mL of 0.60 M MgCl2 c. 200.0 mL of 0.40 M NaC

(SHOW WORK, Expaniton) Thank you.

Answer 1

1.a)

Mass of Sucrose =150 gms

Molecular weight of sucrose (C12H22O11)= 12*12+ 22*1+ 16*11= 144+ 22+ 176 =342

Moles of sucrose =150/342 = 0.438596 gmoles

Molarity is defined as no of moles of solute ( in this case sucrose ) in 1 liter of solution

250 ml is there is 0.439596 gmoles

1000ml is there is 0.439856*1000/250=1.754M

b) .

Volume of methanol =125 ml and density= 0.792 g/ml

Mass of methanol = volume * density= 125 *0.792 gms= 99 gms, moles of methanol (CH3OH)= mass/molecular weight

=99/(12+3+16+1)= 99/32= 3.09375 gmoles

Molar concentration = moles/ volume = 3.09375/15 =0.20625 gmoles/liter

c)

Molarity of NaOH solution =0.4 M =0.4moles/liter

Moles in 250 ml = 0.4*250/1000=0.1 moles

Grams of Sodium hydroxide in 250ml =0.1*molecular weight =0.1*40=4 gms

d)

1ppb= 1mg/1000ml

Lead in 15 mg =15mg/1000L

Atomic weight of lead = 207

Moles of lead in 15mg =15*10-3g/207=7.246*10^-5 moles

Molarity= moles of pb/liter of solution = 7.246X10^-5/1000 M =7.246*10^-8M

e)

Moles of sucrose in 0.423M in 250ml =(250/1000)*0.423=0.10575 moles

Let the volume of sucrose be V

V* 0.650 =0.10575

V=0.10575/0.65 =0.162liter or 162 ml

f)

100ml of 0.3 M AlCl3 corresponds to 0.3*100/1000 =0.03 moles

Moles of Cl- =3*moles of AlCl3= 3*0.03=0.09moles

On similar lines moles of Cl- in MgCl2 = 0.6*(50/1000)*2 =0.06 moles and

Moles of Cl- in NaCl =(200/1000)*0.4 =0.08 moles

AlCl3 contains largest number of chloride ions (0.09 moles