Practice problem G5 [ENDORSED]

[Madysen Ratsek 3L]

I am struggling with problem G5 from the textbook. I think the unit "mmol" is tripping me up slightly and is making me confused for the rest of the problem. Would anyone be able to help walk through it? Particularly the part where we need to know the volume of a particular part of the solution (as in, 2.15 mmol of Na+ instead of considering the whole molecule of Na2CO3. Thanks in advance !!

[Annica Reyes 2B]

Hi, Madysen!

Addressing the unit confusion, I would just convert 2.15 mmol Na+ to 2.15 x 10^-3 mol Na+. To first answer part a of the question, I think we should find the molarity of Na2CO3, which we would do by dividing 2.111g by (105.99g/mol)(.25L). We got .25 L from 250 mL and 105.99 is the molar mass. From this, we can find the volume needed. From before, we multiply 2.15 x 10^-3 mol Na+ with 1 mol of Na2CO3. We then divide that by the molarity of Na2CO3 that we got before multiplied by 2 moles of Na+. As a result, we get .0135 L, or 1.35 x 10^-2 L, the volume of solution needed to transfer to a flask to obtain 2.15 mmol Na+. I hope I covered everything! Let me know if I should tweak anything.

[Kaira Shibata 1E]

I am struggling with problem G5 from the textbook. I think the unit "mmol" is tripping me up slightly and is making me confused for the rest of the problem. Would anyone be able to help walk through it? Particularly the part where we need to know the volume of a particular part of the solution (as in, 2.15 mmol of Na+ instead of considering the whole molecule of Na2CO3. Thanks in advance !!

Hi! I think the mol can be confusing at times. It is always good to have your numbers in the base unit so convert it to moles and then go ahead and work from there. (1000 mmol=1 mol) Also, for the Na+, you just have to look at the formula for the compound. In this case, there are 2 Na+ in one Na2CO3. With this information, you simply multiply the molarity by 2 because it is a 2:1 ratio.

[Madysen Ratsek 3L]

Hi, Madysen!

Addressing the unit confusion, I would just convert 2.15 mmol Na+ to 2.15 x 10^-3 mol Na+. To first answer part a of the question, I think we should find the molarity of Na2CO3, which we would do by dividing 2.111g by (105.99g/mol)(.25L). We got .25 L from 250 mL and 105.99 is the molar mass. From this, we can find the volume needed. From before, we multiply 2.15 x 10^-3 mol Na+ with 1 mol of Na2CO3. We then divide that by the molarity of Na2CO3 that we got before multiplied by 2 moles of Na+. As a result, we get .0135 L, or 1.35 x 10^-2 L, the volume of solution needed to transfer to a flask to obtain 2.15 mmol Na+. I hope I covered everything! Let me know if I should tweak anything.

Thank you so much !! This really helped.

[Madysen Ratsek 3L]

I am struggling with problem G5 from the textbook. I think the unit "mmol" is tripping me up slightly and is making me confused for the rest of the problem. Would anyone be able to help walk through it? Particularly the part where we need to know the volume of a particular part of the solution (as in, 2.15 mmol of Na+ instead of considering the whole molecule of Na2CO3. Thanks in advance !!

Hi! I think the mol can be confusing at times. It is always good to have your numbers in the base unit so convert it to moles and then go ahead and work from there. (1000 mmol=1 mol) Also, for the Na+, you just have to look at the formula for the compound. In this case, there are 2 Na+ in one Na2CO3. With this information, you simply multiply the molarity by 2 because it is a 2:1 ratio.

Thanks! It is easy to get tripped up but this def helps lol. :)

[Iman Gauhar 3E]

I am struggling with problem G5 from the textbook. I think the unit "mmol" is tripping me up slightly and is making me confused for the rest of the problem. Would anyone be able to help walk through it? Particularly the part where we need to know the volume of a particular part of the solution (as in, 2.15 mmol of Na+ instead of considering the whole molecule of Na2CO3. Thanks in advance !!

Hi! I think the mol can be confusing at times. It is always good to have your numbers in the base unit so convert it to moles and then go ahead and work from there. (1000 mmol=1 mol) Also, for the Na+, you just have to look at the formula for the compound. In this case, there are 2 Na+ in one Na2CO3. With this information, you simply multiply the molarity by 2 because it is a 2:1 ratio.

Thanks! I was confused about what to do after finding the molarity, and your explanation was really helpful.

[Annica Reyes 2B]

Hi, Madysen!

Addressing the unit confusion, I would just convert 2.15 mmol Na+ to 2.15 x 10^-3 mol Na+. To first answer part a of the question, I think we should find the molarity of Na2CO3, which we would do by dividing 2.111g by (105.99g/mol)(.25L). We got .25 L from 250 mL and 105.99 is the molar mass. From this, we can find the volume needed. From before, we multiply 2.15 x 10^-3 mol Na+ with 1 mol of Na2CO3. We then divide that by the molarity of Na2CO3 that we got before multiplied by 2 moles of Na+. As a result, we get .0135 L, or 1.35 x 10^-2 L, the volume of solution needed to transfer to a flask to obtain 2.15 mmol Na+. I hope I covered everything! Let me know if I should tweak anything.

Thank you so much !! This really helped.

No problem! I wasn't sure if my explanation was clear or not as I added so many numbers in so many places, but I'm glad I was able to help!