# What is the mass of 5.42 cm^3 of table sugar?

From this site I learnt that the density of table sugar is $700$ $k g \cdot {m}^{-} 3$. So the answer is a bit under $4$ $g$.
In handier units this is: $700 \cdot \cancel{k g} \cdot \cancel{{m}^{-} 3} \times {10}^{3} \cdot g \cdot \cancel{k {g}^{-} 1} \times {10}^{-} 6 \cdot \cancel{{m}^{3}} \cdot c {m}^{-} 3$ $=$ $0.7 \cdot g \cdot c {m}^{-} 3$, or $0.7 \cdot g \cdot m L$.
So (finally) the answer to your question is $5.42 \cdot \cancel{c {m}^{-} 3} \times 0.7 \cdot g \cdot \cancel{c {m}^{3}}$ $=$ $4 \cdot g$ approximately!
It is a fact that sugar DOES NOT dissolve in hexanes. Given that ${\rho}_{\text{hexanes}}$ $=$ $0.67 \cdot g \cdot m L$, should sugar float in hexanes? Can you account for the fact that sucrose, ${C}_{12} {H}_{22} {O}_{11}$ is insoluble in hexanes?