Question

Concentration of pbno3 solution is 0.953m and 0.907m. Calculate density

Answer 1

Answer:

molality = moles solute / kg solvent

So you need to determin the moles solute and mass of solvent from the info provided.

Work with 1 L of solution

molarity = moles solute / litres of solution

So 1 L of a 0.907 M solution has

0.907 moles of Pb(NO3)2

mass Pb(NO3)2 = moles x molar mass

= 0.907 mol x 331.2 g/mol

= 300.3984 g

The mass of the 1 L of solution = density x volume

= 1.252 g/ml x 1000 ml

= 1252 g

Now, mass solvent = total mass - mass solute

= 1252 g - 300.3984 g

= 951.6016 g

= 0.9516 kg

molality = moles / kg

= 0.907 mol / 0.9516 kg

= 0.953 m

Answer 2

Answer:

The density of  PbNO₃ solution is 1.224 g/ml

Explanation:

Given,

The concentration of PbNO₃ solution = 0.953m and 0.907m

Formula:

$m = \frac{M * 1000}{(d * 100) - M * Molecular mass}$     .............................(1)

Where,

• m = molality
• M = Molarity
• d = density

To Find :

Density of PbNO₃ solution :

$d = \frac{\frac{M * 1000}{m} + (M * Molecular mass)} {1000}$    ............................(2)

$Molality = \frac{moles solute}{kg solvent}\\ Molality = 0.953 m\\\\Molarity = \frac{moles of solute}{litres in solution}\\ Molarity = 0.907 m\\\\Molecular mass = moles * molar mass\\ = 0.907 mol * 331.2 g/mol\\Molecular mass = 300.3984 g$

Note : molar mass of PbNO₃ solution  = 331.2 g/mol

Substitute all these values in equation (2) we get the density of PbNO₃ solution.

$d = \frac{\frac{0.907 * 1000}{0.953} + (0.907 * 300.3984)} {1000}$

$d = \frac{1224.19}{1000}\\ d = 1.224 g/ml$

Therefore, the density of  PbNO₃ solution is 1.224 g/ml

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