Question

How many mL of a 0.300 M AgNO3 solution will it take to make 500 mL of a 0.100 M AgNO3 solution?

Answer 1

The answer is 166 mL. You use the formula V1*M1=V2*M2. For what you're given you rearrange the formula to V1=M2*V2/M1 so V1= (0.500 mL)(0.100 M AgNO3)/(0.300 M AgNO3) = 166 mL

Answer 2

Answer: We need 166, 7 mL of AgNO3 [0.300M] to prepare 500mL of AgNO3[ 0.100M]

Step-by-step explanation:

When calculating the amount of volumen from an initial solution in order to prepare other one. we always use this formula for dilutions.

C1V1=C2V2

• First we gather the information that will be replaced in our equation.

Inicial conditions that will be represented by number 1

C1= 0.300M AgNO3

V1= ?

Final conditions that will be represented by number 2

C2= 0.100M AgNO3

V2= 500mL

• Second, we replace the parameters giving in the equation and find out the value for V1.

V1= C2. V2 / C1

V1= 0.100M. 500mL /0.300M

V1= 166.7 mL