Answer
See the answer below.
Work Step by Step
a) Pressure difference between the bottom and the top of the tree:
$10.0\ m_{H_2O}\times 1000\ mmH_2O/m_{H_2O}\times (1\ mmHg/13.6\ mmH_2O)\times 1\ atm/760\ mmHg=0.967\ atm$
The minimum osmotic pressure would need to be greater than this pressure difference for the solute to rise: $\Pi=cRT\rightarrow c=0.967\ atm\div(0.082\ atm/M.K\times 293\ K)=0.04\ M$
b) Mass of 1 L of the solution: $1000\ mL\times 1.0\ g/mL=1000\ g$
Mass of sucrose: $0.04\ mol/L\times 1\ L\times 342.30\ g/mol=13.78\ g$
Weight percent: $13.78/1000\times 100\%=1.38\%$
