Answer
$$ 0.2502 \mathrm{\ m}^{3} $$
Work Step by Step
we konw that
volume of liquid $$"V" =\frac{\rho_{\text {liq. }}}{m_{\text {liq.}}}$$
specific gravity $$"S.G" = \frac{\rho_{\text {liq.}}}{\rho_{H_{2} O}}$$
$$\rho_{\text {liq.}}= S.G. \times \rho_{H_{2} O}$$
"The density of water at $$15^{\circ} \mathrm{C}=999 \mathrm{\ Kg} / \mathrm{m}^{3}, \quad \mathrm{m}_{\text {liq. }}=500 \mathrm{\ Kg}$$
$$\rho_{\text {liq.}}=2 \times 999=1998 \mathrm{\ Kg} / \mathrm{m}^{3}$$
$$V=\frac{1998}{500}=0.2502 \mathrm{\ m}^{3}$$
