Question

Describe, in detail, how to make 3.0 liters of a 3.5 molar HCl solution from an 18.5 molar HCl stock solution

Answer 1

C1V1=C2V2,
where C is concentration and V is volume,
 so let say C1=18.5M, V1=?,
C2=3.5M, V2=3.0L,
 now solve for V1, in this case V1=0.57L, so in order to make a 3.0L solution we will need to take 0.57L of stock HCl and add 2.43L of water. 

Answer 2

Answer:

Take 0.57 L of 18.5 M HCl solution and make its volume upto 3 L by adding it 2.43 L of water.

Explanation:

$M_1V_1=M_2V_2$  (Dilution)

where,

$M_1\text{ and }V_1$ are the , molarity and volume of acid which is formed after dilution

$,M_2\text{ and }V_2$ are the , molarity and volume of acid which is diluted

We are given:

$M_1=3.5 M\\V_1=3.0 L\\M_2=18.5 M\\V_2=?$

Putting values in above equation, we get:

$3.5 M\times 3.0 L=18.5 M\times V_2\\\\V_2=0.57 L$

Volume of 18.5 M HCl required = 0.57 L

Volume to be added in 0.57 L  of 18.5 M HCl so that we have 3.5 M of HCl solution:

$V_1-V_2=3.0 L - 0.57 L = 2.43 L$

Take 0.57 L of 18.5 M HCl solution and make its volume upto 3 L by adding it 2.43 L of water.