There are two aqueous HCl solutions kept in different vessel. How much solution-2 should be added to solution-1 in order to make 2.5M HCl solution?

Question

There are two aqueous HCl solutions kept in different vessel. <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-fjqrofjxhdjvpac7.jpg" /> How much solution-2 should be added to solution-1 in order to make 2.5M HCl solution? (A) 100ml (B) 200ml (C) 500ml (D) 2 litre

Tardigrade Admin · Accepted Answer

2× 500+5× V1=2.5× (500 + V1) 1000+5V1=1250+2.5V1 2.5V1=250 V1=100ml

Q. There are two aqueous $H Cl$ solutions kept in different vessel.

How much solution-2 should be added to solution-1 in order to make $2.5 M$ $H Cl$ solution?

$100 m l$

$200 m l$

$500 m l$

$2$ litre

$2 \times 500 + 5 \times V_{1} = 2.5 \times (500 + V_{1})$
$1000 + 5 V_{1} = 1250 + 2.5 V_{1}$
$2.5 V_{1} = 250$
$V_{1} = 100 m l$

Q. There are two aqueous HCl solutions kept in different vessel. How much solution-2 should be added to solution-1 in order to make 2.5M HCl solution?

100ml

200ml

500ml

2 litre