Question

How do you prepare 250.00 mL of 3.00 M HCl solution starting with a concentrated HCl solution which is 36.0% HCl by mass and has a density of 1.18 g/mL?

Answer 1

Answer: 64.1 ml of concentrated $HCl$ is taken and (250-64.4)= 185.6 ml f water is added to make 250.00 mL of 3.00 M HCl .

Explanation:

Given : 36 g of $HCl$ is dissolved in 100 g of solution.

Density of solution = 1.18 g/ml

Thus volume of solution = $\frac{mass}{density}=\frac{100}{1.18}=84.7ml$

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{36g}{36.5g/mol}=0.99moles$  

$V_s$ = volume of solution

$Molarity=\frac{0.99\times 1000}{84.7}=11.7M$

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock $HCl$ solution = 11.7M

$V_1$ = volume of stock $HCl$ solution = V ml

$M_2$ = molarity of diluted $HCl$ solution = 3.00 M

$V_2$ = volume of diluted $HCl$ solution = 250.0 ml

$11.7M\times V=3.00\times 250.0$

$V=64.1ml$

Thus 64.1 ml of stock $HCl$ is taken and (250-64.4)= 185.6 ml of water is added to make 250.00 mL of 3.00 M HCl .