Commercially available concentrated sulfuric acid is 17.2M . Calculate the volume of concentrated sulfuric acid required to prep
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Answer:
Number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.
Number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.
Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.
Explanation:
1 mole of any compound contains 6.023 × 10²³ molecules.
molecular weight of NaCl is 23 + 35.5 = 58.5 g.
so, 58.5 grams of NaCl makes 1 mole
⇒ 14.5 g of NaCl = = 0.248 moles.
⇒ 0.248 mole contains 0.248 × 6.023×10²³ molecules
= 1.49 × 10²³ molecules.
And 1 molecule contains 1 Na ion and 1 Cl ion.
⇒ number of Na ions in 14.5 g of NaCl is 1.49 × 10²³.
⇒ number of Cl ions in 14.5 g of NaCl is 1.49 × 10²³.
Total number of ions = 1.49 × 10²³ + 1.49 × 10²³ = 2.98 × 10²³.