An adulterated sample of milk has density of $1032\ kg\ m^{- 3}$, while pure milk has a density of $1080\ kg\ m^{- 3}$. Then the volume of pure milk in a sample of 10 L of adulterated milk is

Question

Dr. Abubacker · Accepted Answer

(d) Mass of adulterated milk
$M_{A} = 1032 	imes 10 	imes 10^{- 3}$
$= 10.32\ kg\ \ \ \ \ \ \ \ \ \ \ \ $
Mass of pure milk $\ \ M_{p} = 1080V_{p}$
$	herefore\ \ \ \ \ \ \ Mass\ of\ water\ ho_{w}V_{w} = M_{A} - M_{p}$
$\Longrightarrow \ \ 10^{3}\left( 10 	imes 10^{- 3} - V_{p} ight) = 10.32 - 1080V_{p}$
$\Longrightarrow \ 10 - 10^{3}V_{p} = 10.32 - 1080V_{p}$
$\Longrightarrow \ \ 80V_{p} = 0.32$
$	herefore\ \ V_{p} = \frac{0.32}{80}m^{3}$
$= \frac{0.32}{80} 	imes 1000L = 4L$

Q. An adulterated sample of milk has density of $1032\ kg\ m^{- 3}$, while pure milk has a density of $1080\ kg\ m^{- 3}$. Then the volume of pure milk in a sample of 10 L of adulterated milk is

Explanation

Reviewed by Dr. Abubacker

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