How many grams of FeCl3 are present in 15.00 mL of the solution? (Molar mass of FeCl3 = 162.2 g/mol)
Solution:
V=15.00mL=15.00⋅10−3L;V=15.00mL=15.00\sdot10^{-3}L;V=15.00mL=15.00⋅10−3L;
M(FeCl3)=162.2g/mol;M(FeCl_3)=162.2 g/mol;M(FeCl3)=162.2g/mol;
ν=VVM;\nu=\dfrac{V}{V_{M}};ν=VMV;
ν=15.00⋅10−322.4=0.67⋅10−3mol;\nu=\dfrac{15.00\sdot10^{-3}}{22.4}=0.67\sdot 10^{-3}mol;ν=22.415.00⋅10−3=0.67⋅10−3mol;
ν=mM(FeCl3) ⟹ m=ν⋅M(FeCl3);\nu=\dfrac{m}{M(FeCl_3)}\implies m=\nu\sdot M(FeCl_3);ν=M(FeCl3)m⟹m=ν⋅M(FeCl3);
m=0.67⋅10−3mol⋅162.2g/mol=108.67⋅10−3g=0.109g;m=0.67\sdot10^{-3}mol\sdot162.2g/mol=108.67\sdot 10^{-3}g=0.109g;m=0.67⋅10−3mol⋅162.2g/mol=108.67⋅10−3g=0.109g;
Answer: m=0.109g.m=0.109g.m=0.109g.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments