# Concentration/Mass Question (1 Viewer)

#### _Anonymous

##### Member
Need help with these. If someone could explain (worked solution) the answers to these, it’d be greatly appreciated. Thanks.

1)A beaker contains 100mL of Silver Sulfate solution at a concentration of 1.0 x 10^-6 mol/L.

What is the mass of Silver ions in the beaker?

2) A student adds 200mL of distilled water to 20mL of 4.0 M solution of Sodium Chloride. What is the concentration of the diluted solution of Sodium Chloride?

I did C1V1 = C2V2
4 x 0.02 = C2 x 0.20, therefore C2 = 0.40

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#### fan96

##### 617 pages
1)

Because there are $\bg_white 10^{-6}$ moles of silver sulfate in every litre, to find the number of moles in 100 mL we simply divide by 10.

$\bg_white \mathrm{n(Ag_2SO_4)} = 10^{-6} \times 0.1 =10^{-7}$

For every mole of silver sulfate there are two silver ions.

$\bg_white n(Ag) = 2 \times 10^{-7}$

Multiply by the molar mass of silver to get $\bg_white 215.7364 \times 10^{-7} = 2.2 \times 10^{-5}g \,(2sf)$

That conflicts with the answer you provided. I'm not sure which one is right.

2)

Reread the question, your $\bg_white V_2$ is wrong.

#### _Anonymous

##### Member
1)

Because there are $\bg_white 10^{-6}$ moles of silver sulfate in every litre, to find the number of moles in 100 mL we simply divide by 10.

$\bg_white \mathrm{n(Ag_2SO_4)} = 10^{-6} \times 0.1 =10^{-7}$

For every mole of silver sulfate there are two silver ions.

$\bg_white n(Ag) = 2 \times 10^{-7}$

Multiply by the molar mass of silver to get $\bg_white 215.7364 \times 10^{-7} = 2.2 \times 10^{-5}g \,(2sf)$

That conflicts with the answer you provided. I'm not sure which one is right.

2)

Reread the question, your $\bg_white V_2$ is wrong.
Yeah sorry, I wrote down question 1 incorrectly.

2) Could you elaborate on how my V2 is wrong? I thought that C1V1 is the equation that applies to the substance before it's diluted and C2V2 is after it's diluted?

So how is V2 not 200mL

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#### JaxsenW

##### Active Member
Because the question says that 200ml of distilled water is added to the 20ml solution, your V2 becomes 200+20ml, and that gives you the 0.36
If the question says the solution is diluted to 200ml, then your answer is correct.

Be careful of the wording of the question

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#### _Anonymous

##### Member
Because the question says that 200ml of distilled water is added to the 20ml solution, your V2 becomes 200+20ml, and that gives you the 0.36
If the question says the solution is diluted to 200ml, then your answer is correct.

Be careful of the wording of the question

Sent from my SM-G935F using Tapatalk
Oh, that makes sense; thanks. I thought that adding water would make a solution diluted, hence they would be the same. Could you explain the difference between one adding water and the other being diluted?

#### fan96

##### 617 pages
Oh, that makes sense; thanks. I thought that adding water would make a solution diluted, hence they would be the same. Could you explain the difference between one adding water and the other being diluted?
For aqueous solutions, diluting is the same as adding water.

The expression "adds 200 mL distilled water to 20 mL of solution" is the same as saying "dilutes the solution to 220 mL". But you have to remember the existing 20 mL of solution.

#### _Anonymous

##### Member
For aqueous solutions, diluting is the same as adding water.

The expression "adds 200 mL distilled water to 20 mL of solution" is the same as saying "dilutes the solution to 220 mL". But you have to remember the existing 20 mL of solution.
Ahh right, makes sense. Thanks.