Question

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CALCULATE THE MASS OF A NON-VOLATILE SOLUTE ( MOLAR MASS = 40 gm${mol}^{-1}$ ) WHICH SHOULD BE DISSOLVED IN 114 gm OCTANE TO REDUCE ITS VAPOUR PRESSURE TO 80% .


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Answer 1

$P_{s}$ $= 80\%$

$P_{0} = 0.80$

Solute $= \frac{W}{40g/mol ^{ - 1} }$

Solvent (Octane) $= \frac{114g}{114g/mol ^{ - 1} }$

Octane has formula mass of $114.23g/mol$

$\bold{Carbon}$

$12.0107 \times 8 = 96.0856g$

$\bold{Hydrogen}$

$1.00794 \times 18 = 18.14292g$

$\bold{Total}$

$96.0856 + 18.14292 = 114.22852g/mol$

$\bold{Now,}$

Molar mass of $C_{8}H_{18}$ $= 114g/mol^{ - 1}$

$\bold{Now,}$

$\frac{P_{0} - P_{s} }{P_{0}} = x_{2}$

$\bold{Therefore,}$

$\frac{P_{0} - 0.80P_{0}}{P_{0}} = \frac{ \frac{W}{40} }{ \frac{W}{40} + 1 }$

$\frac{0.8W}{40} = 0.2$

$0.8W = 0.2 \times 40$

$0.8W = 8$

$W = \frac{8}{0.8}$

$W = 10g$