Solution.
C=5.2⋅10−3mol/L;C=5.2\sdot10^{-3} mol/L;C=5.2⋅10−3mol/L;
M(C27H46O)=386.65g/mol;M(C_{27}H_{46}O)=386.65g/mol;M(C27H46O)=386.65g/mol;
V=4.7L;V=4.7L;V=4.7L;
m−?;m-?;m−?;
ν=mM; ⟹ m=ν⋅M;\nu=\dfrac{m}{M};\implies m=\nu\sdot M;ν=Mm;⟹m=ν⋅M;
The molar concentration is calculated as follows:
С=νV ⟹ С=\dfrac{\nu}{V} \impliesС=Vν⟹ ν=C⋅V;\nu=C\sdot V;ν=C⋅V;
ν=5.2⋅10−3mol/L⋅4.7L=24.44⋅10−3mol;\nu = 5.2\sdot10^{-3}mol/L\sdot4.7L=24.44\sdot10^{-3}mol;ν=5.2⋅10−3mol/L⋅4.7L=24.44⋅10−3mol;
m=24.44⋅10−3mol⋅386.65g/mol=9.45g;m=24.44\sdot10^{-3}mol\sdot386.65g/mol=9.45g;m=24.44⋅10−3mol⋅386.65g/mol=9.45g;
Answer: m=9.45g.m=9.45g.m=9.45g.
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