Hello, bosses, fellow developers and programmers, i am aware this isn't the right forum to post this thread; i have been banned from posting on the NL Education forum. I have the following questions from my pre-lab manual. If you can help me with solutions, i will be ever grateful.
.
1. You want to prepare a 50cm3 of 0.1M NaOH from exactly 50% (wt/wt) NaOH solution. What volume of the 50% solution is required?

2. An aqueous sample is analyzed and found to contain 4% NaOH:
.
a. Calculate the molar conc of the solution, assuming the density of water is 1.00g/cm3
.
b. If the solution is neutralized with 1.0M H2SO4, what volume of the acid will be needed?

1. measure ~ 0.19 mL into a 50 mL volumetric flask/container and make it up to the 50 mL mark. ...is it correct?

I did chemistry like 6yrs ago, can't remember shît about it again.

LaiveNg:
2. An aqueous sample is analyzed and found to contain 4% NaOH:
.
a. Calculate the molar conc of the solution, assuming the density of water is 1.00g/cm3
.
b. If the solution is neutralized with 1.0M H2SO4, what volume of the acid will be needed?

I hated this topic

iHart:
1. measure ~ 0.19 mL into a 50 mL volumetric flask/container and make it up to the 50 mL mark. ...is it correct?

Thanks, but i think the questions require a bit math to buttress the significance of the figures.

LaiveNg:

Thanks, but i think the questions require a bit math to buttress the significance of the figures.

but the question was to state the volume and not to show working. so that's why I just gave you that...

LaiveNg:
2. An aqueous sample is analyzed and found to contain 4% NaOH:
.
a. Calculate the molar conc of the solution, assuming the density of water is 1.00g/cm3
.
b. If the solution is neutralized with 1.0M H2SO4, what volume of the acid will be needed?

a. 1.0M

b. 50 mL (assuming the basis of the % is same with the volume of base)

1. you need to convert 50%(wt / wt ) to wt/ v and den find the molar conc.
note: 50g of water will give 50ml/cm3 of water since the density of water is 1.0g/cm3
note also: 1ml =1cm3
to do that (50g of naoh ÷ 50cm3 of diluent/h20) ×100cm3
dat should give u 1g/100cm3 which is the mass conc in 100cm3
but in 1000cm3 the solute will be 10g
therfore
final mass conc is 10g/1000cm3 or 10g/dm3



now convert the mass conc to molar conc
molar conc= mass conc/ molar mass of. naoh
10/40= 0.25M

used this formular RV/O
R= required conc 0.1 M
V= required volume 50ml

0= original conc 0.25Molar

(0.1×50)/0.25
=20ml

therfore 20ml of 50%w/w naoh is needed and made itnup with 80ml.of diluent.

horlahwaley:
1. you need to convert 50%(wt / wt ) to wt/ v and den find the molar conc.
note: 50g of water will give 50ml/cm3 of water since the density of water is 1.0g/cm3
note also: 1ml =1cm3
to do that (50g of naoh ÷ 50cm3 of diluent/h20) ×100cm3
dat should give u 1g/100cm3 which is the mass conc in 100cm3
but in 1000cm3 the solute will be 10g
therfore
final mass conc is 10g/1000cm3 or 10g/dm3



now convert the mass conc to molar conc
molar conc= mass conc/ molar mass of. naoh
10/40= 0.25M

used this formular RV/O
R= required conc 0.1 M
V= required volume 50ml

0= original conc 0.25Molar

(0.1×50)/0.25
=20ml

therfore 20ml of 50%w/w naoh is needed and made itnup with 80ml.of diluent.

WOW!!! Thanks a bunch! You're worth it. Can i please post more

LaiveNg:
Hello, bosses, fellow developers and programmers, i am aware this isn't the right forum to post this thread; i have been banned from posting on the NL Education forum. I have the following questions from my pre-lab manual. If you can help me with solutions, i will be ever grateful.
.
1. You want to prepare a 50cm3 of 0.1M NaOH from exactly 50% (wt/wt) NaOH solution. What volume of the 50% solution is required?

I think the answer should rather be 0.4 mL.

First of all what is the Molar conc of 50% wt/wt of NaOH?
50% wt/wt means 50g NaOH/100g water.
If we assume density of water is 1...
then you have 50g NaOH /100mL water.
thus the mass conc = 50g/0.1L (note that 0.1L = 100 mL).
thuse mass conc = 500g/L.
but molar mass of NaOH is ~ 40g/mol
and more so molar conc = mass conc/ molar mass
therefore the molar conc = 500g/L ÷ 40g/L
= 12.5mol/L or 12.5M
now using C1V1 = C2V2 .....(1)
where C1 is the initial molar conc = 12.5M
V1 is the volume of the initial solution you will need to measure out = unknown.
C2 is the Molar conc of the solution you want to prepare = 0.1 M
V2 is the volume of the solution you want to prepare = 50mL.
thus making V1 subject of the formula from equation (1) you will have ;
V1 = (C2V2)/C1
thus V1 = 0.4 mL.

so just measure out 0.4 ml into a 50mL flask and make up the volume to the mark.

Does this one look correct?

People now bring their assignment to Nairaland everyday . . .

iHart:

Does this one look correct?

Frankly, you won't know how much this has helped me! You demystified it all for me. I have even used this approach to solve other problems in the manual.
.
But i don't understand why we had to covert 100ml to 0.1L.
.
And, i am battling with this also:
.
What amount of oxalic acid (H2C2O4.2H2O, MW = 126.0g/mol) is required to prepare 10cm3 of 0.130M oxalic acid solution?
.
I tried using C1V1 = C2V2, but i can't figure out C1.

LaiveNg:

.
But i don't understand why we had to covert 100ml to 0.1L.

the SI unit of molar conc is mol/dm^3 (note dm^3=L)

LaiveNg:

.
And, i am battling with this also:
.
What amount of oxalic acid (H2C2O4.2H2O, MW = 126.0g/mol) is required to prepare 10cm3 of 0.130M oxalic acid solution?
.
I tried using C1V1 = C2V2, but i can't figure out C1.

you can't use C1V1 = C2V2 because you were not given the conc of the starting material unlike in the first question where you were given 50% wt/wt.


let's take a long route so you can understand it. you want to prepare 10mL (0.01 L) of 0.13 M oxalic acid. what is the mol of what you want prepare?
recall that Molar conc = no of mol/volume(L).
therefore no of mol of what you want to prepare = molar conc x volume(L)
= 0.130 M x 0.01 L
= 0.0013 mol

what amount in gram (mass) of oxalic acid do you need to have in 10 ml (0.01 L) such that the mol will be 0.0013 mol?

this leads us to the relationship of mol and mass.
recall that no of mol = mass(g)/molar mass(g/mol)
therefore the mass will be => no of mol x molar mass(g/mol)
= 0.0013 mol x 126.0 g/mol
= 0.1638 g


thus you need to weigh out 0.1638 g of the oxalic into a 10 ml flask/container and make up the volume to the 10mL mark.

in a nutshell amount (g) you will need = molar conc of what you to prepare x the volume in L of what you want prepare x molar mass of the compound...

Thank you so much boss! I really, really appreciate this... But please help me to understand. Is it that "number of moles" the same thing as "molarity" and "molar concentration"? And if yess, does it mean there are different mathematical relationships for calculating the concentration of a substance? Pardon my ignorance. I had a poor background in Chemistry.

LaiveNg:
Thank you so much boss! I really, really appreciate this... But please help me to understand. Is it that "number of moles" the same thing as "molarity" and "molar concentration"? And if yess, does it mean there are different mathematical relationships for calculating the concentration of a substance? Pardon my ignorance. I had a poor background in Chemistry.

They are not the same.

lolz, this looks like shorthand for my face. my brain no even carry books again but just to count money