Question

A chemist places 2.5316 g of Na 2SO 4 in a 100 mL volumetric flask and adds water to the mark. She then pipets 15 mL of the resulting solution into a 50 mL volumetric flask and adds water to the mark and mixes to make a solution. She then pipets 15 mL of this new solution into a 50 mL volumetric flask and dilutes to the mark. Determine the molar concentration of sodium sulfate in the most dilute solution prepared.

Answer 1

Answer:

The concentration of the most dilute solution is 0.016M.

Explanation:

First, a solution is prepared and then it undergoes two subsequent dilutions. Let us calculate initial concentration:

$[Na_{2}SO_{4}]=\frac{moles(Na_{2}SO_{4})}{liters(solution)} =\frac{mass((Na_{2}SO_{4}))}{molarmass(moles(Na_{2}SO_{4}) \times 0.100L)} =\frac{2.5316g}{142g/mol\times 0.100L } =0.178M$

First dilution

We can use the dilution rule:

C₁ x V₁ = C₂ x V₂

where

Ci are the concentrations

Vi are the volumes

1 and 2 refer to initial and final state, respectively.

In the first dilution,

C₁ = 0.178 M

V₁ = 15 mL

C₂ = unknown

V₂ = 50 mL

Then,

$C_{2}=\frac{C_{1} \times V_{1} }{V_{2}} =\frac{0.178M \times 15mL}{50mL} =0.053M$

Second dilution

C₁ = 0.053 M

V₁ = 15 mL

C₂ = unknown

V₂ = 50 mL

Then,

$C_{2}=\frac{C_{1} \times V_{1} }{V_{2}} =\frac{0.053M \times 15mL}{50mL} =0.016M$