Question

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 1

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 %.