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umang

Solve -

6.9 N KOH solution in water contains 30% by weight of KOH . The density of solution will be ?????

aditya_arora04

Hello, my friend,

Do it as follows :

1. Normality, N = ( W_B * 1000 ) / ( E_B * V_S )

Where symbols have there uaual meanings.

2. N = ( W_B * 1000 ) / ( E_B * [ M / Rho ] ) [ rho = density of solution ]

3. N = ( W_B * 1000 * Rho ) / ( E_B * M ) [ M = Total mass of

solution ]

4. Now, let the mass of solution be 100 gm, so, mass of KOH in solution

is 30 gm.

5. So, put the respective values in the formula and you will get rho.

Hope i am correct. But please do tell me if i am wrong.

snehavenus

hi,

here is an direct formula

normality=(x% * density *10) / equivalent mass.

please do tell me if it is wrong

.............

umang

Hey aditya ,

does E implies for molecular mass ?

Also , why did u multiply the eqn. by 1000 ?

Pls explain soon !!!

snehavenus

hey umang.......according to my knowledge E is equivalent mass and i think the equation is multiplied by 1000 if the volume is given in cm^3.if the volume is given in dm³ then we need not multiply by 1000........

am i correct?

umang

Hey sneha ,

I did this by taking molarity = 6.9M as nf=1 . Thats why i asked if it is molecular mass . And yes , u r right about the volume's units .The correct answer is 1.288

phalgun_g

why do u need to bother

just use this simple formula

normality=(x% * density *10) / equivalent mass

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