The distance between Na^(+) and Cl^(-) ions in solid NaCl of density 43.1 g cm^(-3) is _______________ x 10^(-10) m. (Nearest Integer) (Given : N_(A) = 6.02 x 10^(23) mol^(-1))

rho=fracZ x Ma^3 x N_A 43.1=frac4 x 58.5a^3 x 6.02 x 10^23 a^3=0.9 x 10^-23 \ =9 x 10^-24 a=2.08 x 10^-8 cm =2.08 x 10^-10 m for NaCl, distance between Na^+and Cl^-=(a / 2) =1.04 x 10^-10 m

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Chemistry

Exam

Chemistry

The distance between Na⁺ and Cl^$-$ ions in solid NaCl of density 43.1 g cm^{$-$ 3} is _______________ $\times$ 10^{$-$ 10} m. (Nearest Integer)

(Given : N_A = 6.02 $\times$ 10²³ mol^{$-$ 1})

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Correct Answer:

Explanation

\rho=\frac{Z \times M}{a^{3} \times N_{A}}

43.1=\frac{4 \times 58.5}{a^{3} \times 6.02 \times 10^{23}}

\begin{aligned} & a^{3}=0.9 \times 10^{-23} \\\\ & =9 \times 10^{-24} \end{aligned}

a=2.08 \times 10^{-8} \mathrm{~cm}

=2.08 \times 10^{-10} \mathrm{~m}

for

\mathrm{NaCl}

, distance between

\mathrm{Na}^{+}

and

\mathrm{Cl}^{-}=\frac{a}{2}

=1.04 \times 10^{-10} \mathrm{~m}

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