how to solve this question? is it by taking antilog of -0.784 or something… Also is this level of calculation or even finding log of some number required for imat?

hey,

100/18 is not 0.18 tho? shouldnt it be 1000/18…?

You’re right my bad! i’ll let someone else answer this one

its totally fine we all make mistakes XD

Hi!

we know that pH= -log[H+].

[H+]=log^-pH=10^-0.784=0.16.

Then i used the relation:

M(concentration)=moles solute/volume solution(L)

0.16=n/1

n=0.16.

Then using n=m/Molar mass

m=n x M.M=0.16 X 36.5=6g(approximately).

I had to use calculator to find the value of [H+] also the answer was not precise i just took approximation. Hope IMAT will not ask questions involving calculator.

Hope it helps:)

hey

just to point out another way

if we have most important logarithmic equations memorized, we can conclude that log 6 =0.778========) -log 6^-1 =0.778

then the concentration would be 1/6

one mole is 36.5

36.5/6 approximately 6

thank you!

also( if at all) some log values needs to be memorized its just log 1to10 right

yes

log 1 to 7 would also suffice