1-15 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} 1\ut{u}&=1\ut{g/mol}\\ 1\ut{mol}&=6.022~140~76\times10^{23}\\ m_H&=1.0\ut{u}\\ m_O&=16\ut{u} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} m_{(H_2O)}&=1.0\ut{u}\times2+16\ut{u}\\ &=18\ut{u}\\ &=18\ut{u}\times\frac{1\ut{g}}{1\ut{u}\cdot1\ut{mol}}\times\frac{1\ut{kg}}{1000\ut{g}}\\ &=\frac{18}{6.022~140~76\times10^{23}\cdot1000}\ut{kg}\\ &\approx2.988970320912924\times10^{-26}\ut{kg}\\ &\approx3.0\times10^{-26}\ut{kg}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Ans&=\frac{2.8\times10^8\ut{kg}}{18\ut{u}}\\&=\frac{2.8\times10^8\ut{kg}}{18\ut{u}}\times\frac{1000\ut{g}}{1\ut{kg}}\times\frac{1\ut{u}\cdot1\ut{mol}}{1\ut{g}}\\ &=\frac{2.8\times10^{11}}{18}\times6.022~140~76\times10^{23}\\ &\approx9.367774515555555\times10^{33}\\ &\approx9.4\times10^{33}\\ \end{aligned} $$