When 1.685 g of an alkali metal chloride is | vedclass

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(B) The elevation in boiling point is given by $\Delta T_b = i 	imes K_b 	imes m$. For an alkali metal chloride $(MCl)$,the van't Hoff factor $i = 2

Vedclass · Accepted Answer

$\frac{7}{\sqrt{3}} \ \mathring{A}$

Vedclass · Answer

(B) The elevation in boiling point is given by $\Delta T_b = i 	imes K_b 	imes m$. For an alkali metal chloride $(MCl)$,the van't Hoff factor $i = 2$. Given $\Delta T_b = 100.051 - 100 = 0.051 \ ^\circ C$,$K_b = 0.51 \ K \ kg \ mol^{-1}$,mass of solute $w = 1.685 \ g$,and mass of solvent $W = 200 \ g$. $0.051 = 2 	imes 0.51 	imes \frac{1.685}{M} 	imes \frac{1000}{200}$. $0.051 = 1.02 	imes \frac{1.685}{M} 	imes 5$. $0.051 = \frac{8.425}{M} \implies M = \frac{8.425}{0.051} \approx 165.2$. Considering the given atomic masses,$CsCl$ $(133 + 35.5 = 168.5)$ is the closest match. $CsCl$ crystallizes in a $bcc$ structure where the edge length $a$ is related to the radii by $\frac{\sqrt{3} a}{2} = r^+ + r^-$. Given $r^+ = 1.70 \ \mathring{A}$ and $r^- = 1.80 \ \mathring{A}$,$r^+ + r^- = 3.50 \ \mathring{A}$. $\frac{\sqrt{3} a}{2} = 3.5 \implies a = \frac{7}{\sqrt{3}} \ \mathring{A}$.

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