Question

29.2% (w/w) hcl stock solution has density of 1.25 g ml1. The molecular weight of hcl is 36.5 g mol1. The volume (in ml) of stock solution required to prepare a 200 ml solution of 0.4 m hcl is

Answer 1

Volume (in ml) of stock solution required to prepare a 200 ml solution of 0.4 m HCl is 8.

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

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

where,

n = moles of solute

$V_s$ = volume of solution in ml

Given : 29.2g of HCl in 100 g of solution.

Volume of solution =$\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.25g/ml}=80ml$

moles of HCl= $\frac{\text {given Mass}}{\text {Molar Mass}}=\frac{29.2g}{36.5g/mol}=0.8$

Now put all the given values in the formula of molarity, we get

$Molality=\frac{0.8\times 1000}{80ml}=10M$

To calculate the volume of stock solution

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are molarity and volume of stock solution.

$M_2\text{ and }V_2$ are molarity and volume of resulting solution

We are given:

$M_1=10M\\V_1=?mL\\\M_2=0.4M\\V_2=200mL$

Putting values in above equation, we get:

$10\times V_1=0.4\times 200\\\\V_1=8mL$

Thus the volume (in ml) of stock solution required to prepare a 200 ml solution of 0.4 m HCl is 8.

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Answer 2

8 mL

Let us assume the mass of the stock of solution = 100 grams

Mass of HCl in 100 g of 29.2 % (wight by weight) HCl stock solution = 29.2 g

rams

Volume of the stock solution:

=  Mass Of Solution / Density

= 100g/1.25 g mL

= 80 grams

Number of moles of HCl in the stock solution:

= Given Mass / Molar Mass

= 100 / 1.25 = 0.8

Molality of the stock solution = (0.82/80)×1000

= 10 M

Now using the formula,

M1V1 = M2V2

10 x V1 = 0.4 x 200

10 x V1 = 80

V1 = 80 / 10

V1 = 8 mL

Therefore, the volume required is 8 mL.