2 chem 12 questions - need help!

1 Calculate the percent by mass of acetic acid, CH3COOH, in a solution of 0.620M CH3COOH. Assume the density of the solution is 1.07 g/mL

#2 How many moles of KOH(s) must be added to 200 mL of a 0.100M solution of HNO3 to bring the pH to 2.00

If you could do a quick step by step for each question it would be greatly appreciated.

1.

With 1 liter CH3COOH , moles of CH3COOH = x = 0.62 mol
⇒ the mass of CH3COOH = m = 0.62 × 60 = 37.2 g
and d = 1.07g/ml
⇒ the mass of 1 liter CH3COOH = 1 000 ml is: 1.07 × 1000 = 1070 g
⇒ %(mCH3COOH) = 37.2 / 1070 = 3.4766 %
2.
V = 200ml = 0.2 litters (const)

moles of HNO3 = Σn(H+) = 0.1× 0.2 = 0.02 mol

moles of KOH = a = Σn(OH–)
we have:
H(+) + OH (–) –-–-–-–->H2O

pH = 2 ⇔ [H+ residual] = 10^-2 = 0.01 M
⇔ n(H+ residual ) / V = 0.01
⇔ n(H+residual) = 0.01 × 0.2 = 0.002 mol
⇔ Σn(H+) – Σn(OH–) = 0.002
⇔ 0.02 – a = 0.002
⇔ a = 0.018 mol

1.we set the volume of the solution is V(unit:L),so the total mass of the solution is 1.07×1000V(because 1L=1000mL),the acid n=0.62×V(because M=mol L-1) hence the mass of acid m=0.62×V×60
then the percent by mass of the acid
w=m(acid)/m(solution)
=0.62*60*V/(1.07*1000*V)
=3.48%
2.pH=2.00 means [H+]=10^(-2.00)=0.01M
in 0.100M solution of HNO3 [H+]=0.1M
we set x mol KOH must be added ,KOH+HNO3=KNO3+H2O,[H+]=(0.1M×0.2L-x mol)/0.2L=0.01M
we get x=0.018 mol