Posted by **Jason** on Sunday, November 7, 2010 at 10:48am.

Plese tell me if these are right.

A 15.0-mL sample of NaCl solution has a mass of 15.78g. After the NaCl is evaporated to dryness, the dry salt residue has a mass of 3.26g. Calculate the following concentrations for the NaCl solution.

a. %(m/m)= 3.26 x 100 / 15.78 = 20.7

b. %(m/v) = 3.26 x 100 / 15.0 = 21.7

c. molarity (m) Moles NaCl = 3.26 g / 58.4428 g/mol = 0.0558

M = 0.0558 mol / 0.0150 L = 3.72

How many grams of KI are in 25.0 mL of a 3.0%(m/v) KI solution. 3.0 = mass KI x 100 / 25.0

mass KI = 0.750 g

How many milliliters of a 2.5 M MgCl2 solution contain 17.5g Mg Cl2. Moles MgCl2 = 17.5 g / 95.211 g/mol= 0.184

V = 0.184 mol / 2.5 M = 0.0736 L => 73.9 mL

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