A50.0 −ml sample of 1.50×10−2 m na2so4(aq) is added to 50.0 ml of 1.28×10−2 m ca(no3)2(aq). what percentage of the ca2+ remains unprecipitated?

When we have Ksp CaSO4 = 9.1x10^-6 so by substitution in Ksp formula:
Ksp CaSO4 = [Ca+2][SO4] 
1.9x10^-6   = [Ca+2][1.5x10^-2]
∴ [Ca+2] M in the solution =1.27x10^-4
when percent remaining = [Ca in solution]/[C original value] * 100
                                          = [(1.27x10^-4*(50ml+50ml)]/[1.28x10^-2*50ml]*100
                                          ≈ 2%

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml

Ithink it is quantity of plants