Question

29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. The molecular weight of HCl is 36.5 g mol-1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is..​

Answer 1

Answer:

8mL

Explanation:

Step1. found the molarity from the given condition

Step 2. By using M1V1=M2V2, found the volume

Answer 2

hey mate,

your question

29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. The molecular weight of HCl is 36.5 g mol-1. The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is..

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

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}Molarity=

V

s

n×1000

where,

n = moles of solute

V_sV

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

density of solution

mass of solution

=

1.25g/ml

100g

=80ml

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

Molar Mass

given Mass

=

36.5g/mol

29.2g

=0.8

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

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

80ml

0.8×1000

=10M

To calculate the volume of stock solution

M_1V_1=M_2V_2M

1

V

1

=M

2

V

2

where,

M_1\text{ and }V_1M

1

and V

1

are molarity and volume of stock solution.

M_2\text{ and }V_2M

2

and V

2

are molarity and volume of resulting solution

We are given:

Putting values in above equation, we get:

\begin{lgathered}10\times V_1=0.4\times 200\\\\V_1=8mL\end{lgathered}

10×V

1

=0.4×200

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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