Question

CHEM

Calculate the molarity (M) ov the following solutions
a. 2.0 moles of glucose in 4.0L of solution
b. 4.0 g of KOH in 2.0 L of solution
c. 5.85 g NaCl in 400. mL of solution ...?

Answer 1

Molarity =moles of solute /volume of solution. A.) Molarity of glucose =moles of glucose /volume of solution =2/4 = 0.5 M B.) No of moles of KOH= weight of KOH / molecular weight of KOH = 4/ 56.10 =0.07 moles. Therefore, Molarity of KOH =moles of KOH /volume of solution = 0.07/2 = 0.035M. C.) No of moles of NaCl = weight of NaCl/ molecular weight of NaCl = 5.85/58.5 = 0.1moles. Therefore, Molarity of NaCl = moles of NaCl /volume of solution = 0.1/ 400 x 10^-3 = 0.25M

Answer 2

Molarity = moles / liter

a) M = 2/4 = 0.5 M

b) Moles = 4/(30 + 16 + 1)
= 0.085
M = 0.085 / 2 = 0.0425 M

c) Moles = 5.85 / (23 + 35.5)
= 0.1
M = 0.1 / 0.4
= 0.25 M