Compute the number of ions present in 5.85 g of sodium chloride?
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Mol. wtofNaCl= 23+35.5=58.5 g
58.5 g of Na Cl = 1 mole
5.85 g ofNaCl = 1 X 5.85/58.5 = 0.1 mole
Each molecule of NaCI contains one Na ion and one Cl ion = 2 ions
Total moles of ions in 0.1 mole of NaCI = 2 x 0.1 = 0.2 moles
No. of ions = 0.2 x 6.022 x 1023 = 1.2044 x 10^23 ions
58.5 g of Na Cl = 1 mole
5.85 g ofNaCl = 1 X 5.85/58.5 = 0.1 mole
Each molecule of NaCI contains one Na ion and one Cl ion = 2 ions
Total moles of ions in 0.1 mole of NaCI = 2 x 0.1 = 0.2 moles
No. of ions = 0.2 x 6.022 x 1023 = 1.2044 x 10^23 ions
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