Question

10 ml of N/2 HCL, 20 ml of M / 5 H2SO4 , 20 ml of N/4 H3PO4 are mixed together. the volume of water that must be added to make a 0.18 N solution is? ( assume there is no change in volume on mixing). Ans.50

EXPLAINATION REQUIRED!!!!!!!

Answer 1

Answer:

We know that Meq = N × V

Now, Meq of HCl = 1/2 × 10 = 5

Similarly, Meq of H3PO4 = 1/4 × 20 = 5

For H2SO4, N = 2M

So, Meq of H2SO4 = 2/5 × 20 = 8

So, combined Meq = 5+5+8 = 18

If final volume is 100ml

Then resulting concentration = 18/100 = 0.18N

Now we already have 10ml HCl + 20ml H2SO4 + 20ml H3PO4 = 50ml

Required volume of water = 100-50 = 50ml

I hope you get the point.

Answer 2

The volume of water that must be added is 50ml.

Given:-

Volume of HCl = 10ml

Volume of H2SO4 = 20ml

Volume of H3PO4 = 20ml

Normality of HCl = N/2

Morality of H2SO4 = M/5

Normality of H3PO4 = N/4

Normality of Solution (n) = 0.18N

To Find:-

The volume of water that must be added.

Solution:-

We can simply calculate the volume of water that must be added by using the following process.

As

Volume of HCl (v1) = 10ml

Volume of H2SO4 (v2) = 20ml

Volume of H3PO4 (v3) = 20ml

Normality of HCl (N1) = N/2

Morality of H2SO4 (M) = M/5

Normality of H3PO4 (N3) = N/4

Normality of Solution (N) = 0.18N

Volume of solution = v =?

Normality of H2SO4 = Molarity * n-factor

N2 = M/5*2 =2M/5

Now, according to the conservation of moles,

$n = n1 + n2 + n3$

$n = n1 + n2 + n3$$</strong></p><p><strong>[tex]n = n1 + n2 + n3$$N \times v = </strong></p><p><strong>[tex]n = n1 + n2 + n3$$N \times v = N1v1 + </strong></p><p><strong>[tex]n = n1 + n2 + n3$$N \times v = N1v1 + N2v2 + </strong></p><p><strong>[tex]n = n1 + n2 + n3$$N \times v = N1v1 + N2v2 + N3v3$

$n = n1 + n2 + n3$$N \times v = N1v1 + N2v2 + N3v3$$0.18n \times v = ( \frac{n}{2} \times 10) + ( \frac{2n}{5} \times 20) + ( \frac{n}{4} \times 20)$

on canceling, n from both the sides,

[tex]0.18 \times v = \frac{10}{2} + \frac{2}{5} \times 20 + \frac{20}{4} [/tex]

tex]0.18 \times v = \frac{10}{2} + \frac{2}{5} \times 20 + \frac{20}{4} [/tex]$0.18 \times v = 5 + 8 + 5$

tex]0.18 \times v = \frac{10}{2} + \frac{2}{5} \times 20 + \frac{20}{4} [/tex]$0.18 \times v = 5 + 8 + 5$$0.18 \times v = 18$

tex]0.18 \times v = \frac{10}{2} + \frac{2}{5} \times 20 + \frac{20}{4} [/tex]$0.18 \times v = 5 + 8 + 5$$0.18 \times v = 18$$v = \frac{18}{0.18}$

tex]0.18 \times v = \frac{10}{2} + \frac{2}{5} \times 20 + \frac{20}{4} [/tex]$0.18 \times v = 5 + 8 + 5$$0.18 \times v = 18$$v = \frac{18}{0.18}$$v = \frac{1800}{18}$

tex]0.18 \times v = \frac{10}{2} + \frac{2}{5} \times 20 + \frac{20}{4} [/tex]$0.18 \times v = 5 + 8 + 5$$0.18 \times v = 18$$v = \frac{18}{0.18}$$v = \frac{1800}{18}$$v = 100ml$

tex]0.18 \times v = \frac{10}{2} + \frac{2}{5} \times 20 + \frac{20}{4} [/tex]$0.18 \times v = 5 + 8 + 5$$0.18 \times v = 18$$v = \frac{18}{0.18}$$v = \frac{1800}{18}$$v = 100ml$Now, Volume of water added = 100 - (10+20+20)

tex]0.18 \times v = \frac{10}{2} + \frac{2}{5} \times 20 + \frac{20}{4} [/tex]$0.18 \times v = 5 + 8 + 5$$0.18 \times v = 18$$v = \frac{18}{0.18}$$v = \frac{1800}{18}$$v = 100ml$Now, Volume of water added = 100 - (10+20+20)v' = 100 - 50 = 50ml

Hence, The volume of water that must be added is 50ml.

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