1-32 할리데이 11판 솔루션 일반물리학

풀이에 필요한 정보가 너무 부족하여 고양이의 질량과 구성성분비는 임의로 정했습니다. $$\begin{cases} \text{Cat} &= 5\ut{kg}\\ 1\ut{mol}&=6.02\times10^{23}\\ 1\ut{u}&=1\ut{g/mol} \end{cases}$$ $$\begin{cases} O : 70\ut{\%}\\ C : 20\ut{\%}\\ H : 10\ut{\%}\\ \end{cases}$$ $$\begin{cases} m_O &= 16\ut{u}\\ m_C &= 12\ut{u}\\ m_H &= 1.0\ut{u}\\ \end{cases}$$ $$\begin{cases} M_O &= 5\ut{kg}\cdot0.7=3.5\ut{kg}\\ M_C &= 5\ut{kg}\cdot0.2=1\ut{kg}\\ M_H &= 5\ut{kg}\cdot0.1=0.5\ut{kg}\\ \end{cases}$$ $$\begin{aligned} n_O &= \frac{M_O}{m_O}\\ &=\frac{3.5\ut{kg}}{16\ut{u}}\cdot\frac{1\ut{u}\cdot1\ut{mol}}{1\ut{g}}\cdot\frac{1000\ut{g}}{1\ut{kg}}\\ &=\frac{875}{4}\ut{mol}&\cdots(1) \end{aligned}$$ $$\begin{aligned} n_C &= \frac{M_C}{m_C}\\ &=\frac{1\ut{kg}}{12\ut{u}}\cdot\frac{1\ut{u}\cdot1\ut{mol}}{1\ut{g}}\cdot\frac{1000\ut{g}}{1\ut{kg}}\\ &=\frac{250}{3}\ut{mol}&\cdots(2) \end{aligned}$$ $$\begin{aligned} n_H &= \frac{M_H}{m_H}\\ &=\frac{0.5\ut{kg}}{1.0\ut{u}}\cdot\frac{1\ut{u}\cdot1\ut{mol}}{1\ut{g}}\cdot\frac{1000\ut{g}}{1\ut{kg}}\\ &=500\ut{mol}&\cdots(3) \end{aligned}$$ $$(1),(2),(3)$$ $$\begin{aligned} \Ans&=\Sigma n\\ &=\frac{875}{4}\ut{mol}+\frac{250}{3}\ut{mol}+500\ut{mol}\\ &=\frac{9625}{12}\ut{mol}\\ &\approx802.0833333333334\ut{mol}\\ &\approx8.0\times10^2\ut{mol}\\ \end{aligned}$$