1. How many moles of CH3COOH would be required to produce a 2.5 M solution with

Question

1. How many moles of CH3COOH would be required to produce a 2.5 M solution with a volume of 1.5 L?
2. How many grams of CCl4 would be used to produce a 2.0 M solution with a volume of 1.00 L?
3. What is the molarity of a 2.50 x 10^2 cm^3 solution containing 45.0 g of C6H12O6?
4. How many moles of Ch3CH2-OH would be used in the preparation of 2.50 L of a 3.5 M solution?
If you could show the work it would be so appreciated! 1. How many moles of CH3COOH would be required to produce a 2.5 M solution with a volume of 1.5 L?
2. How many grams of CCl4 would be used to produce a 2.0 M solution with a volume of 1.00 L?
3. What is the molarity of a 2.50 x 10^2 cm^3 solution containing 45.0 g of C6H12O6?
4. How many moles of Ch3CH2-OH would be used in the preparation of 2.50 L of a 3.5 M solution?
If you could show the work it would be so appreciated!
2. How many grams of CCl4 would be used to produce a 2.0 M solution with a volume of 1.00 L?
3. What is the molarity of a 2.50 x 10^2 cm^3 solution containing 45.0 g of C6H12O6?
4. How many moles of Ch3CH2-OH would be used in the preparation of 2.50 L of a 3.5 M solution?
If you could show the work it would be so appreciated!

Explanation / Answer

1. For one liter it need 2.5 moles CH3COOH and for 1.5 L = 2.5 +1.25 (or 2.5 x 1.5) = 3.75 Moles need

2. Molar mass of CCL4 is 153.82. 2M solution for one Liter = 2 (Mole) x 153.82 (Molar mass) = 307.64 gm of CCL4 need

3. 2.50 x 102 cm3 = 2.5 x 102 ml = 2.5 x100 = 250 ml

Molar mass of C6H12O6 is 180.16.

45 gm in 250 ml is equal to 180 gm (45 x 4) in one liter which is 1M solution

Concentration is 1 M

4. 3.5 moles need for One liter hence for 2.5 liter = 3.5 x 2.5 = 8.75 Moles need

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