What is the percent cscl by mass in a 0.711 m aqueous cscl solution that has a density of 1.091 g/ml?

Answer is:  the percent caesium chloride by mass is 11%.
V(CsCl) = 1 L · 1000 mL/L = 1000 mL.
c(CsCl) = 0.711 M.
n(CsCl) = V(CsCl) · c(CsCl).
n(CsCl) = 1 L · 0.711 mol/L.
n(CsCl) = 0.711 mol.
m(CsCl) = n(CsCl) · M(CsCl).
m(CsCl) = 0.711 mol · 168.36 g/mol.
m(CsCl) = 119.7 g.
d(CsCl) = 1.091 g/mL.
mr(CsCl) = d(CsCl) · V(CsCl).
mr(CsCl) = 1091 g.
w/w = 119.7 g ÷ 1091 g × 100%.
w/w = 11 %.

This is what i found on google “when one phosphate group is removed by breaking a phosphoanhydride bond in a process called hydrolysis, energy is released, and atp is converted to adenosine diphosphate (adp). likewise, energy is also released when a phosphate is removed from adp to form adenosine monophosphate (amp).” i hope this you .