The concentration of lead nitrate (Pb(NO₃)₂))in a 0.626 M solution is __________ molal. The density of the solution is 1.202 g/mL. A)0.521 B)1.83 C)0.629 D)0.819 E)0.650

Question

Quizplus · Accepted Answer

To calculate the molality (m), we need to use the formula:
m = moles of solute / mass of solvent in kg

First, let's calculate the moles of solute:
mol Pb(NO3)2 = M x V (where M is the molarity and V is the volume in L)
mol Pb(NO3)2 = 0.626 M x 1 L = 0.626 mol

Next, let's calculate the mass of solvent:
mass of solvent = volume x density
mass of solvent = 1 L x 1.202 g/mL = 1.202 kg

Now we can calculate the molality:
m = 0.626 mol / 1.202 kg = 0.629 mol/kg

Therefore, the concentration of lead nitrate in a 0.626 M solution is 0.629 molal. The answer is (C).

The Concentration of Lead Nitrate (Pb(NO3)2))in a 0

The Concentration of Lead Nitrate (Pb(NO₃)₂))in a 0