Question

A)A volume of 46.2 mL of a 0.568 M Ca(NO₃)₂ solution is mixed with 75.5 mL of a
1.496 M Ca(NO₃)₂ solution. Calculate the concentration of the final solution.

Calculate the molarity of each of the following solutions.

(b) 7.25 g of methanol (CH₃OH) in ×  10² mL of solution:

mol/L

(c) 9.68 g of calcium chloride (CaCl₂) in 2.20 × 10² mL of solution:

mol/L

(d) 8.57 g of naphthalene (C₁₀H₈) in 85.2 mL of benzene solution:

mol/L

Answer 1

ar% орстот (A) M3 (V, + V2) = M,V, + M₂ V₂ M = Mivit M V2 (V, +V₂) 0.568 x 46-2) + (1.496 X 75,5 139.1896 121.7 (46.2 + 75.5) - 26. 2416 + 112.948 121.7 Final M₂ = 1.1437 M Concentration?

(b) Malarity = Mass of Methanol & Molar mass of methanal 1ooo Volume of Solution in mL 000 M = 7.25 x 10 = 3264 C M M = 2.225 M (molle). Mass of Cach Molar mass Volume (2.2 x 102) M - 9.68 x 1000 = 9680 Х І ось 224.4 24908.4 M= 0.388 mol/L

Mass of Cotts x Molar mass toon Volume of benzene M = 8.57 x looo 8570 128 85 - logo5.6 M = 0.785 mol/L

(A) Final concentration of Ca(NO3)2 solution is M3= 1.1437 mol/Liter.

(b) Molarity of methanol solution is M = 2.225 mol/L

(c) Molarity of CaCl2 solution in Water is M = 0.388 mol/L

(d) Molarity of naphthalene in benzene is M = 0.785 mol/L