G.5- Mixtures and Solutions

[Riya Sood 4G]

G. 5) 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 Na1; (b) 4.98 mmol CO3
(c) 50.0 mg Na2CO3?

How do we solve this problem?

[NRobbins_1K]

First, we find out how many moles of sodium carbonate are in 2.111 g so that we can find the molarity of the sodium carbonate solution. We divide the mass by sodium carbonate's molecular weight 2.111/106 to find that we have 0.01992 moles of sodium carbonate. To find the molarity, we divide the moles by the volume in liters, 0.01992m/.25 L to find that the initial solution had a molarity of 0.07967 moles/liter. Problem a) asks to find out what volume of that .0797 M solution would contain 2.15 mol Na. When we look at the sodium carbonate molecular formula, we see that for every mole of Na2CO3, there are two moles of Na. So we can calculate using the formula M=m/V to calculate how what volume it would require to contain 2.15/2= 1.075 mmoles of Na2CO3, and that volume will contain twice as many moles of Na: 2.15 mmol. We use the M=m/V equation to find that it would require 0.0135 liters or 13.5 mL of the solution to contain 2.15 mmol Na. For part b), the procedure is similar, except that there is only one mole of CO3 per mole of sodium carbonate, so we input 4.98 mmol into the M=m/V equation instead of half that amount. Using this, we find that it would require 0.0625 L or 62.5 mL of solution to contain 4.98 mmol CO3. For part c), we must first find out how many moles of sodium carbonate are in 50mg before we perform the same calculation we have been doing. We simply divide 50mg by the molar mass of sodium carbonate, 106, to find that 50mg represents 4.71698113×10−4 moles of Na2CO3. We simply input this molar amount into the formula M=m/V once again to find that it would require 0.005921 liters or 5.921 mL of solution to contain 50mg of sodium carbonate.

[Mariepahos4D]

A) convert from moles grams of sodium carbonate to moles to get .0199 mols Na2CO3. Get the molarity of the solution by dividing the moles by the liters of solution so .0199 mols/.25L= .0796 M
A wants you to find the volume in a solution containing 2.5mmol of Na. 1 mol contains 10^3 mmol so 2.15mmol/10^3= .00215 mols Na. To get .00215 mols of Na you need half as much Na2CO3 since the subscipt of the Na is 2 meaning that for every one mol of Na2CO3 there are two mols of Na.
So .00215/2= .001075 mols Na2CO3 needed.
Using the molarity .0796M= .001075mols/ V to get V= .0135L or 13.5 ML required

B) 4.98 mmol .00498 mols CO3 so similar to part A .0796M= .00498mols/ V = .063L or 62.6 Ml

C) convert the mg to g and then to moles using molar mass to get 5.29955 mols Na2CO3. Then use the molarity and moles to solve for volume. .0796=5.29955/V= 66.6L

[Mariepahos4D]

First, we find out how many moles of sodium carbonate are in 2.111 g so that we can find the molarity of the sodium carbonate solution. We divide the mass by sodium carbonate's molecular weight 2.111/106 to find that we have 0.01992 moles of sodium carbonate. To find the molarity, we divide the moles by the volume in liters, 0.01992m/.25 L to find that the initial solution had a molarity of 0.07967 moles/liter. Problem a) asks to find out what volume of that .0797 M solution would contain 2.15 mol Na. When we look at the sodium carbonate molecular formula, we see that for every mole of Na2CO3, there are two moles of Na. So we can calculate using the formula M=m/V to calculate how what volume it would require to contain 2.15/2= 1.075 mmoles of Na2CO3, and that volume will contain twice as many moles of Na: 2.15 mmol. We use the M=m/V equation to find that it would require 0.0135 liters or 13.5 mL of the solution to contain 2.15 mmol Na. For part b), the procedure is similar, except that there is only one mole of CO3 per mole of sodium carbonate, so we input 4.98 mmol into the M=m/V equation instead of half that amount. Using this, we find that it would require 0.0625 L or 62.5 mL of solution to contain 4.98 mmol CO3. For part c), we must first find out how many moles of sodium carbonate are in 50mg before we perform the same calculation we have been doing. We simply divide 50mg by the molar mass of sodium carbonate, 106, to find that 50mg represents 4.71698113×10−4 moles of Na2CO3. We simply input this molar amount into the formula M=m/V once again to find that it would require 0.005921 liters or 5.921 mL of solution to contain 50mg of sodium carbonate.

For part C, you first have to convert mg to grams before using the molar mass to convert to moles as molar masses are represented in grams.

[405745446]

For part C, after finding out Molarity which the posts above have shown:
convert 50mg (Na2CO3) to 0.05 g

Then divide the grams by molar mass to get moles :
0.05g /105.99g*mol^-1 = 4.717x10^-4

Plug values into M=m/v equation:
0.0796M = 4.717x10^-4/ v

Get v by itself:
v = 4.717x10^-4mol/0.0796M
= 59.259x10^-4L

Convert L to mL
5.9259x10^-3 *1000mL/1L = 5.926mL should be the answer in the book.

[Aaron Kwan 3B]

Make sure to keep track of units (mg, g, L, m)!