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The number of unit cells in 58.5 g of NaCl is nearly ........(A) 6 \(\times\) 1020(B) 3 \(\times\) 1022(C) 1.5 \(\times\)1023(D) 0.5 \(\times\) 1024 |
Answer» Correct answer is: (C) 1.5 \(\times\) 1023 58.5 g of NaCl = 1 mole = 6.023 \(\times\) 1023 NaCl units. One unit cell contains 4 NaCl units. Hence, the number of unit cells present = \(\cfrac{6.023\times10^{23}}{4}=1.5\times10^{23}\) |
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