Question

Larutan jenuh air kapur mengandung 0,148 Ca(OH)2 (Mr = 74) per liter. Jika massa jenis larutan tersebut adalah 1,050 kg/liter. Hitunglah konsentrasi larutan tersebut dalam a) M b) N c) m d) X dan e) % berat

Answer 1

Dik:v=1.050 kg/liter
Dit:a.M
b.N
c.m
d.x
C.%berat
Jawab:
a.M=n/v
=14.18/1.050
=13.50 m

b.N Ca(OH)=1.050/74
=14.18 mol

c.m=gr/mr×1000/gr pelarut
=1050/74×1000/0.418
=1050/10.952
=95.87 molar

E.%Ca=m×Ar/mr Ca(OH)2×100%
=1×40/40+32+2×100%
=40/74×100%
=40/74×100%-54.05%
%=2×16/74×100%
=32/74×100%
=43.24%

%H=2×1/74×100%
=2/74×100%
=2.70%

D.n Ca(OH)2=1050/74
=14.18mol

n Ca=1/1×14.18mol
=14.18mol

n H=1/1×14.18mol
=14.18 mol