Question

Kian is working in the lab with a 9.50M stock solution. If he makes a solution with a volume of 1.25L and a molarity of 3.00M, what was the volume of his stock solution?

Answer 1

Answer :

The stock solution occupies 3.96 L

Given :

Kian prepares a 3.00 M solution from a 9.50 M stock solution which occupies 1.25 L

Explanation :

We have to find the volume occupied by the 3.00 M stock solution

Dilution Equation

$\large{\star \ \boxed{\boxed{\sf M_1V_1 = M_2V_2 }}}$

$\sf{Here}\begin{cases} \sf{M_1,M_2 \longrightarrow Molarity} \\ \sf{V_1,V_2 \longrightarrow Volume }\end{cases}$

Substituting the values,

$\longrightarrow \: \sf \: 9.50 \times 1.25 = 3.00 \times v \\ \\ \longrightarrow \: \sf \: 3v = 11.875 \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: v = 3.96 \: L}}$

Answer 2

...

$\tt{\huge{\purple{ .................. }}}$

QuEsTi0N -

Kian is working in the lab with a 9.50M stock solution.

If he makes a solution with a volume of 1.25L and a molarity of 3.00M, what was the volume of his stock solution ?

S0LUTI0N -

This question can be solved very easily by using the Dilution Equation.

• Dilution Equation -

• $\sf{\boxed{\boxed{\orange{ \leadsto{\huge{ C_{1} V_{1} = C_{2} V_{2} }}}}}}$

Where -

C_{1} and C_{2} are the respective concentrations .

V_{1} and V_{2} are the respective volumes.

Here -

C_{1} = 9.50 M

C_{2} = 3.00 M

V_{1} = ?

V_{2} = 1.25 L

Substuting the required values in the above Equation -

We obtain -

V_{1} = 3.96 litres

____________________________________