Question

The density of water is 1.00 g/cm3. What is its density in kg/m3?

Answer 1

Answer:

$10^{3}\; \rm kg \cdot m^{-3}$.

Explanation:

Convert the unit in two steps:

• Convert $\rm g$ to $\rm kg$ given that $10^{3}\; \rm g = 1\; \rm kg$.
• Convert $\rm cm^{3}$ to $\rm m^{3}$ given that $10^{6}\; \rm cm^{3} = (10^{2})^{3}\; \rm cm^{3} = 1\; \rm m^{3}$.

Start by converting $\rm g$ to $\rm kg$. Since $10^{3}\; \rm g = 1\; \rm kg$, multiplying the original value by the ratio $\displaystyle \frac{1\; \rm kg}{10^{3}\; \rm g}$ would not change of the value expression:

$\begin{aligned}& 1.00\; \rm \frac{g}{cm^{3}} \\ =\; & 1.00\; \frac{\rm g}{\rm cm^{3}} \times \frac{1\; \rm kg}{10^{3}\; \rm g} \\ =\; & 1.00\times 10^{-3}\; \frac{\rm kg}{\rm cm^{3}}\end{aligned}$.

Convert $\rm cm^{3}$ to $\rm m^{3}$. Notice that the original unit $\rm cm^{3}\!\!$ is in the denominator. Thus, choose a ratio where $\rm cm^{3}\!$ is in the numerator: $\displaystyle \frac{10^{6}\; \rm cm^{3}}{1\; \rm m^{3}}$.

$\begin{aligned}& 1.00\times 10^{-3}\; \frac{\rm kg}{\rm cm^{3}} \\=\; & 1.00\times 10^{-3}\; \frac{\rm kg}{\rm cm^{3}} \times \frac{10^{6}\; \rm cm^{3}}{1\; \rm m^{3}} \\ =\; & 1.00 \times 10^{3}\; \frac{\rm kg}{\rm m^{3}}\end{aligned}$.

In other words, $1.00\; \rm g \cdot cm^{-3} = 1.00 \; \rm kg \cdot m^{-3}$.

Answer 2

Answer:

The metric system was originally devised so that water would have a density of 1 g/cm3, equivalent to 103 kg/m3.