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The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, nonelectrolyte solid weighing \(0.5 \mathrm{~g}\) is added to 39.0 \(\mathrm{g}\) of benzene (molar mass \(78 \mathrm{~g} / \mathrm{mol}\) ). The vapour pressure of the solution then is 0.845 bar. What is the molecular mass of the solid substance?
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The correct answer is:
170
\(P_A^{\circ}=0.850\) bar, \(P_S=0.845\) bar
\(w=0.5 \mathrm{~g}, \quad m=\text { ? }\)
weight of solvent (benzene) \(=39.0 \mathrm{~g}\)
and molecular weight of benzene \(=78 \mathrm{~g}\)
As we know, \(\frac{P_A^{\circ}-P_S}{P_A^{\circ}}=x_B=\frac{n_B}{n_A}\)
\(\frac{0.850-0.845}{0.850}=\frac{\frac{w_B}{M_B}}{\frac{w_A}{M_A}}=\frac{\frac{0.5}{m}}{\frac{39}{78}}\)
On solving, we get molecular mass of solid structure \((\mathrm{m})=170 \mathrm{~g}\).
\(w=0.5 \mathrm{~g}, \quad m=\text { ? }\)
weight of solvent (benzene) \(=39.0 \mathrm{~g}\)
and molecular weight of benzene \(=78 \mathrm{~g}\)
As we know, \(\frac{P_A^{\circ}-P_S}{P_A^{\circ}}=x_B=\frac{n_B}{n_A}\)
\(\frac{0.850-0.845}{0.850}=\frac{\frac{w_B}{M_B}}{\frac{w_A}{M_A}}=\frac{\frac{0.5}{m}}{\frac{39}{78}}\)
On solving, we get molecular mass of solid structure \((\mathrm{m})=170 \mathrm{~g}\).
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