Pharmacy school maths Help

[waxen fractal]

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[obtuse flume]

this isnt the only question i have alot but its just confusing my brain

[dawn sluice]

have you seen the concentration to volume formulae?

[obtuse flume]

Conc = moles/volume

[dawn sluice]

oh no

[weary agate]

nope

[dawn sluice]

i mean $C_1\cdot V_1=C_2\cdot V_2$

[shell field]

temu mod | promote to mod pls

[obtuse flume]

Ohhh yes they always talk about this equation but i never know when to use it

[dawn sluice]

here is a good time, since you have the concentration

and volume

[obtuse flume]

So they concentration is 0.05 g per 100 ml

which is 0.0005 ?

[dawn sluice]

yeah.

[weary agate]

Yes

[dawn sluice]

let me double check

yep!

[weary agate]

pretty sure hes right

or she

[obtuse flume]

so then i do 0.0005 x 1500 = 10 x C1

[obtuse flume]

he!

[obtuse flume]

hold on

so c1 = 0.075

[dawn sluice]

k

nice

[dawn sluice]

now you can use this

[obtuse flume]

why?

[dawn sluice]

you have the concentration, and the volume

[weary agate]

because you have all the stuff

concetration and vol

[dawn sluice]

now you can calculate the moles, then the grams from the periodic table

[obtuse flume]

moles = conc x volume

[weary agate]

then your good

yep

[dawn sluice]

yeap!

[weary agate]

put all the values in there and you should get the answer

[obtuse flume]

so 0,075 x 450,000 = 33750

[dawn sluice]

wha

[weary agate]

not rly

[dawn sluice]

[obtuse flume]

ummm

[weary agate]

Hint?

[obtuse flume]

mhm

[weary agate]

Moles=0.075mol/L×450L

[obtuse flume]

ohh so conc is always in litres right?

[dawn sluice]

no.

[obtuse flume]

oh

[dawn sluice]

volume is liters

[weary agate]

💀

[dawn sluice]

concentration is your percentage

so percentage times liters

[obtuse flume]

:0

[weary agate]

Sry thought you would understand

[obtuse flume]

Dont worry im a bit slow when it comes to math 😭

[obtuse flume]

0.075 x 450

[dawn sluice]

it’s good to practice dimensional analysis when it comes to these kinds of problems

,calc 0.075*450

[shell field]

Result:

33.75

[weary agate]

Yep!

[dawn sluice]

so that’s your moles

[weary agate]

Then

[dawn sluice]

now just use the one given on the periodic table

[weary agate]

?

its 33.75 times _____

[dawn sluice]

specifically, the atomic mass, because that’s your g/mol

[obtuse flume]

we dont get that

for this question

[weary agate]

do you want to know the second number

[dawn sluice]

[obtuse flume]

its 1000

[weary agate]

no

[obtuse flume]

oh

what is it?

because the answer should be 33750

[weary agate]

so there potassium

[dawn sluice]

why would you not be allowed to use the periodic table?

[weary agate]

39.10g/mol

[dawn sluice]

i mean, i can use rough estimations to get the molar mass.

[weary agate]

theres

[obtuse flume]

these questions dont need a periodic table

[weary agate]

54.94g/mol

[obtuse flume]

i dont think they want us to use the molar mass

[dawn sluice]

interesting.

[weary agate]

and oxygen

16.00g/mol×4=64.00g/mol

[dawn sluice]

what is the method, then?

now i’m confused.

[weary agate]

wait what

this is a different problem then

[obtuse flume]

im pretty sure the c1 v1 you mentioned is correct and then i get confused from there but you multiply it by 450,000 to find out how much there is in 450 L

[dawn sluice]

?!?!?

what the hell...

[obtuse flume]

i know ....

[weary agate]

bruh

[dawn sluice]

wouldn’t 450000L=450kL?

[weary agate]

yes

[obtuse flume]

its 450L, so 450,000 Ml

srry i didnt write ml

[weary agate]

yes

[obtuse flume]

but

[dawn sluice]

that makes more sense now...

[obtuse flume]

concentration x volume = mass???

[weary agate]

ohhhh

i got 33750

[dawn sluice]

that’s density

[obtuse flume]

thats right

can you explain plss

i understand up to the 0.075

[weary agate]

is it because they dont want us to use molar mass?

i didnt use molar mass and got 33750

[obtuse flume]

yepp

[weary agate]

so first part your rigt

450 l to ml

450000 ml

Use 0.05%w/v

Mass of KMnO 4 = 0.05 × 450,000 / 100 =22,500g

Then we take the stock solution dilution

[obtuse flume]

😭

[weary agate]

its just chem

the stock solution must be 150 times more concentrated than the working solution

[obtuse flume]

yepp i get that

[weary agate]

because the ratio of dilution is 1:150

[obtuse flume]

Mhm

[weary agate]

SO

Mass of stock solution = 22,500g × 150 = 3,375,000g.

[obtuse flume]

im confused on how you got the 22.5k

so the 0.05 is the concentration

[weary agate]

here

[obtuse flume]

then you multiplied it by the volume?

why u did thats

[weary agate]

the volume of stock solution (450 L) determines how much KMnO 4 is needed for dilution into the working solution. The stock solution is more concentrated, so we account for the 1:150 factor to find the total mass.

[obtuse flume]

Yepp

[weary agate]

got it?

im top yapper today

[obtuse flume]

This message yes

[weary agate]

what else are u confused about

[obtuse flume]

sorry if u are busy 😭

[weary agate]

im not

what else are u confused about

[obtuse flume]

So i understood the dilution factor is 1:150 and the total volume is 450 L

then we have the conc with is 0.05% w/v

[weary agate]

yep

[obtuse flume]

so that means it would be 0.75 in 1500

Now im stuck from this stage onwards

[weary agate]

so after you multiply 0.05 and 450000, you divide all that by 100

and you get 22500 g

[obtuse flume]

ohhhh

yep

[weary agate]

and then

This is a 1:150 dilution.

so

The stock solution must be 150 times more concentrated than the working solution.

[obtuse flume]

mhm

[weary agate]

22500 g x 150 = 3375000 g

[obtuse flume]

then you divide by 100 again?

because the ans is 33750

[weary agate]

ig

yeah

[obtuse flume]

😭 no idea why but ima just do it

i kinda get it

still a bit tricky

but tyvm

[arctic mirage]

@obtuse flume

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[distant badge]

Not sure what they mean by "% w/v". Doesn't make sense to apply percentages to non-dimensionless units. I guess they mean it as A(KMnO4) = 5*10^(-4) g/ml?
For brevity, let X ≡ KMnO4.
Suppose the mass concentrations of X in the stock and working solution are A0(X) and A1(X), respectively. Then we have:
A0(X) V0(X sol.) = A1(X) V1(X sol.)
From here:
A0(X) = A1(X) V1(X sol.)/V0(X sol.)
And for our solution:
A0(X) = m(X)/V(X sol.)
So:
m(X) = A0(X) V(X sol.) = A1(X) V(X sol.) V1(X sol.)/V0(X sol.)
Make sure to convert V(X sol.) to ml first.