Find the volume needed to prepare a 0.694 m pb(no3)2 solution if you start with 110 g of pb(no3)2

0.479 dm^3 (or 479 cm^3)

Explanation:

First, calculate the no. of moles in 110g of Pb(NO3)2.

According to the periodic table, molar mass of Pb is 207.2, N is 14,0, O is 16.0

so, the molar mass of Pb(NO3)2 = 207.2 + (14.0 + 16.0 x 3) x 2 = 331.2

With the formula:

no. of moles = mass/molar mass

no. of moles in 110g of Pb(NO3)2 = 110 / 331.2

= 0.33213 mol

Then, with the formula:

molarity = no. of moles / volume

0.694 = 0.33213 / v

v = 0.479 dm^3 (or 479 cm^3)

Therefore, the volume needed is 0.479 dm^3 (or 479 cm^3)