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 gramsB) 33.1 gramsC) 269 gramsD) 53.8 grams

To calculate the mass of lead nitrate (Pb(NO3)2) needed to make a 1.00 molar solution in 100 milliliters of solution, you can use the formula:

Mass (in grams)
=
Molarity (in moles/liter)
×
Volume (in liters)
×
Molar mass (in g/mol)
Mass (in grams)=Molarity (in moles/liter)×Volume (in liters)×Molar mass (in g/mol)

First, you need to convert the volume from milliliters to liters, which is
100
mL
=
0.1
L
100 mL=0.1 L.

Next, you have the molarity (1.00 M) and the molar mass of lead nitrate (Pb(NO3)2) is given as 331 g/mol.

Now, plug these values into the formula:

Mass (in grams)
=
1.00
mol/L
×
0.1
L
×
331
g/mol
Mass (in grams)=1.00 mol/L×0.1 L×331 g/mol

Calculate the mass:

Mass (in grams)
=
1.00
×
0.1
×
331
=
33.1
grams
Mass (in grams)=1.00×0.1×331=33.1 grams

So, the correct answer is B) 33.1 grams.