A solid having density of 9 times 10^3 kg m | vedclass

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(C) The formula for density is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Given: $d = 9 	imes 10^3 \ kg \ m^{-3}$,$Z = 4$ (for $FCC$),$N_A = 6 	imes 10

Vedclass · Accepted Answer

$0.0305 \ kg \ mol^{-1}$

Vedclass · Answer

(C) The formula for density is $d = \frac{Z 	imes M}{N_A 	imes a^3}$.Given: $d = 9 	imes 10^3 \ kg \ m^{-3}$,$Z = 4$ (for $FCC$),$N_A = 6 	imes 10^{23} \ mol^{-1}$,and $a = 200 \sqrt{2} \ pm = 200 \sqrt{2} 	imes 10^{-12} \ m$.Substituting the values:$9 	imes 10^3 = \frac{4 	imes M}{6 	imes 10^{23} 	imes (200 \sqrt{2} 	imes 10^{-12})^3}$.$a^3 = (200 \sqrt{2} 	imes 10^{-12})^3 = (2 \sqrt{2} 	imes 10^{-10})^3 = 8 	imes 2 \sqrt{2} 	imes 10^{-30} = 16 \sqrt{2} 	imes 10^{-30} \ m^3$.Using $\sqrt{2} \approx 1.414$,$a^3 \approx 16 	imes 1.414 	imes 10^{-30} \approx 22.624 	imes 10^{-30} \ m^3$.$M = \frac{9 	imes 10^3 	imes 6 	imes 10^{23} 	imes 22.624 	imes 10^{-30}}{4} = \frac{54 	imes 22.624 	imes 10^{-4}}{4} = 13.5 	imes 22.624 	imes 10^{-4} \approx 305.4 	imes 10^{-4} \ kg \ mol^{-1} = 0.0305 \ kg \ mol^{-1}$.

$A$ solid having density of $9 \times 10^3 \ kg \ m^{-3}$ forms face-centred cubic crystals of edge length $200 \sqrt{2} \ pm$. What is the molar mass of the solid? [Avogadro constant $\cong 6 \times 10^{23} \ mol^{-1}$,$\pi \cong 3$]

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