The volume of 0.5 M aqueous NaOH solution required to neutralize 10 ml of 2 M aqueous HCL solution is
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we know that
M1V1= M2V2
So,
0.5×V1=2×10
=) V1=40 ml
we can also use N1V1=N2V2
SO,
WE KNOW
N= M×n where n is the n factor
u can use this formula for all cases
M1V1= M2V2
So,
0.5×V1=2×10
=) V1=40 ml
we can also use N1V1=N2V2
SO,
WE KNOW
N= M×n where n is the n factor
u can use this formula for all cases
Ajayprajapati111:
right answer bro . but m1v1= m2v2 is not valid for all reactions . it's better if you go through no. of moles required
so what would be the formula used if we go for no of moles?
let i give you a question just replace NaOH by MgOH2 then . And i according to your method answer is same but that's not the right answer so go and find answer to these question then you understand why i am advicing you .
*know
I hve understood ur point. in general what would be the formula?
we can also use normality for this.
no. of moles = mv that remains the basic formula . no more comments
then it would be N1V1=N2V2 which is equivalent as M1 × n× V1= M2×n×V2
ok. thanks bro
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