-3
$\begingroup$

Find the equivalent weight of an acid if it is given that $\pu{1.321 g}$ acid reacts with magnesium to give $\pu{1.72 g}$ salt. Options:

  1. $49$
  2. $36.5$
  3. $45$
  4. none of above.

The second choice is given as the answer.

I tried to solve the question. I used the formula

$$\frac{\text{wt. of acid}}{\text{equivalent weight of hydrogen} \text{ eq. wt. of anion say 'x'}} = \frac{\text{wt. of salt}}{\text{eq. wt. of magnesium eq. wt. of anion}}$$

And after solving I got $x=35.41$ taking equivalent weight of hydrogen as $1$, and $x=35.38$ taking equivalent weight of H as $1.008$. The answer is different in each case. Is this correct? But the correct answer given is $36.5$

$\endgroup$
  • $\begingroup$ If known salt is given I can find the solution. But I have no idea about magnesium. We can equate no. of gm. eq. wt. of metal = no. of gm. eq. wt. of salt. $\endgroup$ – Sangam Jun 7 '15 at 2:25
  • $\begingroup$ I mean we have weight of metal divided by equivalent weight of metal = wt. of salt divided by equivalent weight of salt. If known salt is given (e.g. MgCl or MgSO4) we can find equivalent weight of salt. And hence by solving the above equation we know the ew of metal. But in this question ew of salt is not given and I dont kmow that how Mg helps to find ew of metal. We also dont have weight of Mg. I need weight of salt or Mg. I am preparing for entrance exam and this is an objective question. 4 options are given: 1)49 2) 36.6 3)45 4) none of above. In answer option b is given. $\endgroup$ – Sangam Jun 7 '15 at 2:48
1
$\begingroup$

The trick is to recognize that the acid and the salt share an anion $\ce{A}$, for example:

$$\ce{2HA + Mg -> MgA2 + H2}$$

Thus the equivalent weights of the acid and the salt are functions of the only one unknown, the equivalent weight of $\ce{A}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.