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
EPORT FOR EXPERIMENT 15 (continued) NAME 3. Calculate the moles and grams of lead(II) nitrate pre...
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