Question #12550
1) what volume of 0.125M CaCl2 can be made from 15.0g
Expert's answer
The molar mass of CaCl2 is
M(CaCl2) = 40 + 2*35.5 = 40 + 71 = 111
g/mol
The amount of CaCl2 is
n(CaCl2) = m(CaCl2) / M(CaCl2) = 15 g / 111 =
0.135 mol
Molarity is
C(CaCl2) = n(CaCl2) / V(solution)
thus,
V(solution) = n(CaCl2) / C(CaCl2) = 0.135 mol / 0.125 M = 1.08 L
M(CaCl2) = 40 + 2*35.5 = 40 + 71 = 111
g/mol
The amount of CaCl2 is
n(CaCl2) = m(CaCl2) / M(CaCl2) = 15 g / 111 =
0.135 mol
Molarity is
C(CaCl2) = n(CaCl2) / V(solution)
thus,
V(solution) = n(CaCl2) / C(CaCl2) = 0.135 mol / 0.125 M = 1.08 L
