There are two sample of $H C l$ having molarity $1 M$ and $0.25 M$ . Find the volume of these sample taken respectively in order to prepare $0.75 M H C l$ solution. (Assume no water is used)

20 m l, 10 m l

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100 m l, 60 m l

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40 m l, 20 m l

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50 m l, 35 m l

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Solution

The correct options are
A $100 m l, 60 m l$
B $20 m l, 10 m l$
C $50 m l, 35 m l$
D $40 m l, 20 m l$
The expression for final molarity is $\frac{V_{1} \times 1 + V_{2} \times 0.25}{(V_{1} + V_{2})} = 0.75 M$

Hence, $0.75 (V_{1} + V_{2}) = V_{1} + V_{2} \times 0.25$

$0.75 V_{1} + 0.75 V_{2} = V_{1} + V_{2} \times 0.25$

$0.5 V_{2} = 0.25 V_{1}$

$\frac{V_{1}}{V_{2}} = 2$

Thus, all the four options are correct answers.

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There are two sample of HCl having molarity 1M and 0.25M. Find the volume of these sample taken respectively in order to prepare 0.75MHCl solution. (Assume no water is used)

The correct options are A 100ml,60ml B 20ml,10ml C 50ml,35ml D 40ml,20mlThe expression for final molarity is V1×1+V2×0.25(V1+V2)=0.75MHence, 0.75(V1+V2)=V1+V2×0.250.75V1+0.75V2=V1+V2×0.250.5V2=0.25V1V1V2=2Thus, all the four options are correct answers.

There are two sample of $H C l$ having molarity $1 M$ and $0.25 M$ . Find the volume of these sample taken respectively in order to prepare $0.75 M H C l$ solution. (Assume no water is used)