Question

The recommended procedure for preparing a very dilute solution is not to weigh out a very small mass or measure a very small volume of stock solution. Instead, it is done by a series of dilutions. A sample of 0.6597 g of KMnO4 was dissolved in water and made up to the volume in a 500.0−mL volumetric flask. A 2.000−mL sample of this solution was transferred to a 1000−mL volumetric flask and diluted to the mark with water. Next, 10.00 mL of the diluted solution was transferred to a 250.0−mL flask and diluted to the mark with water. (a) Calculate the concentration (in molarity) of the final solution. Enter your answer in scientific notation.

Answer 1

Answer: $6.4\times 10^{-7}M$.

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

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.6597g}{158g/mol}=0.004mole$  

$V_s$ = volume of solution in ml = 500 ml

$Molarity=\frac{0.004\times 1000}{500}=0.008M$

According to the dilution law:

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock  solution = 0.008 M

$V_1$ = volume of stock solution = 2 ml

$M_2$ = molarity of resulting solution = ?

$V_2$ = volume of resulting solution = 1000 ml

$0.008\times 2 ml=M_2\times 1000ml$

$M_2=0.000016M$

According to the dilution law:

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock  solution = 0.000016 M

$V_1$ = volume of stock solution = 10 ml

$M_2$ = molarity of resulting solution = ?

$V_2$ = volume of resulting solution = 250 ml

$0.000016\times 10 ml=M_2\times 250ml$

$M_2=0.00000064M=6.4\times 10^{-7}M$

Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

Therefore, the concentration of of the final solution is $6.4\times 10^{-7}M$.