Question

EPORT FOR EXPERIMENT 15 (continued) NAME 3. Calculate the moles and grams of lead(II) nitrate pre solution you used. 0 grams of lead(II) nitrate present in the 10.0 mL of 0.50 M Pb(NO) mol Would the 10.0 mL of 0.50 M Pb(NO3), be sufficient to precipitate the chromate in 0.850 g of sodium chromate? Show supporting calculations and explain. Potassium 5. Calculate the moles of chromate ion in the potassium chromate you used and in the lead(ID chromate obtained. Theoretically, why should these two values agree? answer 1B moles Cro, in K, Cro

Answer 1

3) 0.0050 moles and 1.656 g of lead(ii) nitrate

Explanation:

Volume of lead nitrate given = 10ml = 0.01L

Molarity of lead nitrate given = 0.50 M or 0.5 mol/L

As Molarity = no. Of moles of solute/ volume of solution (in L)

So, no. Of moles of solute = Molarity× volume

No. Of moles of lead nitrate = (0.50 mol L^{​​​​​​-1} )× .01L

= 0.0050 moles lead(ii) nitrate

Molar mass of lead(ii) nitrate = 331.2 g/mol

No. of moles = 0.0050 moles

No. of moles = grams of that substance/molar mass

Or

grams of substance = no.of moles × molar mass of substance

So, grams of lead(ii) nitrate = (0.0050 moles )× (331.2 g/mol)

= 1.656 g lead (ii) nitrate