Question

What volume of 12 M and 4 M HCl must be mixed to prepare 1000 mL of 6 M HCl?​

Answer 1

Answer:

CONSIDERING VOLUME OF 12M AS X AND VOLUME OF 4M SOLUTION AS 1000-X. JUST PUT THESE VALUES IN THE FORMULA AND YOU WILL GET THE ANSWER 250 AND 750 ML RESPECTIVELY.

Answer 2

Answer:

To prepare the solution of 1000ml of 6M of HCl, the volume of 250ml of 12M and 750ml of 4M of HCl must be mixed.

Explanation:

Consider that the V₁ is the volume of 1st solution and V₂ is the volume of 2nd solution.

Given, the total volume of the solution after mixing is equal to 1 Liter.

∴   $V_{1}+V_{2}=1$

    $V_{2}=1-V_{1}$                                                            ...............(1)

Given, the molarity of first solution, $M_{1}=12M$

Molarity of the second solution $M_{2}=4M$

The number of moles final solution (n) will be the sum of the number of moles of 1st (n₁) and 2nd solution (n₂).

$n=n_{1}+n_{2}$                                                               ................(2)

We know that, $Molarity = \frac{n}{V}$

∴   $n=MV$

Put the value of n in eq.(2):

$MV=M_{1}V_{1}+M_{2}V_{2}$

$(6)(1)=(12)(V_{1})+(4)(V_{2})$

$6=12V_{1}+4(1-V_{1})$

$6=12V_{1}+4-4V_{1}$

$8V_{1}=2$

$V_{1}=0.25L$

$V_{1}=250ml$

Put the value of in eq.(1)

$V_{2}=1-0.25\\V_{2}=0.75\\V_{2}=750ml$

Therefore, the volume of V₁ equals to 250ml and the volume of V₂ equals to 750ml should be mixed to prepare 100ml of 6M HCl.