Question

What mass of lead nitrate i.e. $Pb{(N{O_3})_2}$ (formula weight = 331), is needed to make 100 millimetres of a 1.00 molar solution?
A) 438 grams
B) 33.1 grams
C) 269 grams
D) 53.8 grams

Answer 1

In the question, molarity of the lead nitrate solution i.e. 1 is given. Recall the formula to calculate the molarity of the solution. In the formula of molarity, volume must be in litres, but we give volume in millilitres. You also need to find out the moles of lead nitrate solution. Given that molecular mass of lead nitrate is 331 g.

Complete Solution :
- We are given 1.00 molar solution of lead nitrate i.e. molarity of the solution is given.
Molarity of a solution, denoted by M, is defined as the number of moles of the solute present in 1 litre of the solution. Thus, formula for calculating molarity is:
Molarity (M) = $\dfrac{{{\text{Number of moles of solute}}}}{{{\text{Volume of solution in litres}}}}$

- We are also given, volume of the solution = 100 millilitres.
100 mL = 0.1 litres.
- We are also required to calculate the number of moles of solute that is, $Pb{(N{O_3})_2}$ here to use in the formula of molarity.
Formula for calculating number of moles is:
Number of moles = $\dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}}$

- Now, let the mass of $Pb{(N{O_3})_2}$ required to make 100 millimetres of a 1.00 molar solution is $x$.
And, molar mass of $Pb{(N{O_3})_2}$ = 331 grams (given).
Thus, number of moles of lead nitrate, $Pb{(N{O_3})_2}$= $\dfrac{x}{{331}}$
Now, molarity of $Pb{(N{O_3})_2}$ = $\dfrac{{{\text{Moles of }}Pb{{(N{O_3})}_2}}}{{{\text{Volume of the solution (in L)}}}}$
$\Rightarrow 1.00 = \dfrac{x}{{331 \times 0.1}}$
$\Rightarrow x = 33.1$ grams.
Hence, mass of lead nitrate i.e. $Pb{(N{O_3})_2}$ needed to make 100 millimetres of a 1.00 molar solution is 33.1 grams.
So, the correct answer is “Option B”.

Note: You can also be given molality which is denoted by m. It is defined as the number of moles of solute present in 1 kg of solvent. Thus,
Molality (m) = $\dfrac{{{\text{No}}{\text{. of moles of solute}}}}{{{\text{Mass of solvent (in kg)}}}}$