Chem 14a Problem G.5

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Nadeen 3D
Posts: 21
Joined: Fri Sep 28, 2018 12:26 am

Chem 14a Problem G.5

Postby Nadeen 3D » Fri Oct 05, 2018 1:12 am

A student prepared a solution of sodium carbonate by adding 2.111 g of the solid to a 250.0-mL volumetric flask and adding water to the mark. Some of this solution was transferred to a buret. What volume of solution should the student transfer into a flask to obtain (a) 2.15 mmol Na ; (b) 4.98 mmol CO32 ; (c) 50.0 mg Na2CO3?

Can someone please help me start this problem? I am a bit lost because of the word and would like help in the introduction of the problem?

2c_britneyly
Posts: 62
Joined: Fri Sep 28, 2018 12:16 am

Re: Chem 14a Problem G.5

Postby 2c_britneyly » Fri Oct 05, 2018 1:37 am

Hi Nadeen!

You can solve these by using the equation M(initial)*V(initial)=M(final)*V(final). This problem is essentially asking you to find the initial volume used to make these dilutions. First, you want to determine what values you are already given. You are given the final number of moles, which is the same thing as multiplying the final molarity by the final volume. Therefore, the number of moles given to you in parts a,b,and c should replace the right side of the equation. Just don't forget to convert each given value to moles since they have given you mmol and grams. Then, you can determine the initial molarity by converting 2.111g to moles and dividing by .25 L (250mL). After plugging in that number, you solve for initial volume.

Hope this helps

Hovik Mike Mkryan 2I
Posts: 95
Joined: Fri Sep 28, 2018 12:25 am

Re: Chem 14a Problem G.5

Postby Hovik Mike Mkryan 2I » Fri Oct 05, 2018 1:40 am

Hello,
For this equation knowing M(molarity)=mol/ volume (L) is the first step. With the given information, converting the 2.111g Na2CO3 to moles and dividing by 0.250 to get the molarity(0.07967) will help to answer the rest of the question. For a) convert 2.15mmol to mol with 1000mmol=1mol and divide that number by the molarity number X2 to find the amount of volume needed in liters. Dividing by the double molarity is due to there being 2Na atoms in the compound. I hope this helped.


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