Question

A sample of sea water contains $2.8\%$ $NaCl$ by mass and has a density of $1.03$ g/ml. The amount of water that should be evaporated from ${10}^6$ litres sea water in order to have a saturated solution of $NaCl$ is $x\times {10}^2$. The value of $x$ is (solubility of $NaCl$ being $318.83$ g/L pure water):

Answer 1

Volume of sea water $={10}^6$ L $={10}^9$ mL

$\therefore$ Mass of sea water $={10}^9\times 1.03$ g

$\therefore$ Mass of NaCl $=$ $\frac{2.8\times 1.03\times {10}^9}{100}=2.884\times {10}^7$ g

$\therefore$ Mass of pure water $=$ $103\times {10}^7-2.884\times {10}^7$ $=100.116\times {10}^7$ g

$\because$ Solubility in $1$ L water or $1000$ g water is $318.83$ g.

$\because 318.83$ g $NaCl$ is present in $1000$ g $H_{2}O$.

$\therefore 2.884\times {10}^7$ g $NaCl$ is present in $\frac{1000\times 2.884\times {10}^7}{318.83}=9.046\times {10}^7$ g $H_{2}O$

$\therefore$ Pure water which should be evaporated $=$ $100.116\times {10}^7-9.046\times {10}^7$ $=91.07\times {10}^7$g
$=91.07\times {10}^7$ mL $=9107\times {10}^2$ L (density of pure water is 1g/ml)
So, the value of $x$ is $9107$.