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

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

The answer is 3,5g co

The answer options is d