Question

3. Calculate the molarity of a solution made by adding 75.3 g of Na2CO, to a flask and dissolving it with water to create a total volume of 2.75 L. 4. What mass of Kal (in grams) do you need to make 400.0 mL of a 1.65 M KC solution? 5. How many grams of sucrose ( CH2011) are in 2.45 L of 0.525 M sucrose solution? 6. How many mL of a 0.250 M KCl solution contains 34.5 g KCI? 7. To what volume (in mL) should you dilute 100.0 mL of a 2.50 M CaCl2 solution to obtain a 0.550 M Cach solution?

Answer 1

Given mass of Na2CO3 = 75.3g , Volume =2.75 L.

we know that Molarity = mass in grams(x) 1000 Molecular weight“ Vol.of solution(ml) 75.3 g 1.5.98 g/mol * 2.75 1. 2.752 = 1.713 mol/L

Given Molarity of KCl = 1.65 M, the volume of solution = 400 ml Molecular weight = 74.55g/mol.

we know that Molarity = mass in grams(x) 1000 Molecular weight“ Vol.of solution(ml) 1.65mol/L = X(g) 74.55 g/mol 74.589) mot * 1000 400 ml 1.65 * 74.55 * 400 X(g) = 1000 = 49.203 g

We have Molarity of sucrose = 0.525 M , Volume = 2.45 L

we know that Molarity = mass in grams(x) 1000 Molecular weight“ Vol.of solution(ml) 0.525mol/L= X(g) 342.29g/mol 1 . * 2.45 T X(g) = 0.525 * 342.29 * 2.45 = 440.27 g