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Tamiku [17]
3 years ago
14

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 resu

lting 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.
Chemistry
1 answer:
Lera25 [3.4K]3 years ago
3 0

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

<u>First dilution</u>

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

<u>Second dilution</u>

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

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4 0
2 years ago
A solution is prepared by dissolving 15.0 g of NH3 in 250.0 g of water. The density of the resulting solution is 0.974 g/mL. The
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[NH₃]  → 3.24 M  

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Our solute: Ammonia

Our solvent: Water

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3 years ago
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