Answer the question on a separate sheet of paper bearing your mane and date Avog

Question

Answer the question on a separate sheet of paper bearing your mane and date Avogadro's number, N, is 6.023 times 10^23 particles per mol Na = 23.0; O = 16.0; Cl = 35.5; N = 14.0; H = 1.0; s =32.0; K = 39.1; C = 12.1; Cu = 63.5; Fe = 55.8. Calculate: The number of atoms in 4.50 of sodium The number of molecules in 12.0 g of oxygen gas, O_2 The number of formula units of NaCI in a 7.365 g sample of the compound. Calculate the molecular mass (in amu) and molar masses (in g/mol) for. NH_3 SO_2 N_2O_5 Calculate the molar masses (in g/mol) for: KCI Na_2CO_3 CuSO_4^-5H_2O Calculate: the moles of iron in a 6.345 g sample of the metal the mass of 0.450 mol of C_2H_6O.

Explanation / Answer

1. i) 4.5 g g Na = 4.5/23 = 0.195 moles = 0.195 * 6.023*1023 atoms = 1.18*1023 atoms

ii. 12 g O2 = 12/32 moles O2 = (12/32) * 6.023*1023 atoms = 2.25 *1023 atoms

iii. molar mass of NaCl = 23 + 35.5 =58.5 g

7.365 g NaCl = 7.365/58.5 moles = (7.365/58.5) * 6.023*1023 atoms = 7.58 *1022 atoms

No of formula units = 7.58 *1022

2.i. Molecular mass , NH3 = 14 amu + 3*1 amu = 17 amu

Molar mass, NH3 = 14g/mol + 3*1 g/mol = 17 g/mol

ii. SO2, molecular mass = 32 + 2*16 = 64 amu

molar mass = 64 g/mol

iii. N2O5, molecular mass = 2*14 + 5*16 = 108 amu

molar mass = 108 g/mol

3.i. molar mass,

KCl = 39.1 + 35.5 = 74.6 g/mol

Na2CO3 = 23*2 + 12 + 16*3 = 106 g/mol

CuSO4.5H2O = 63.5 + 32 + 4*16 + 5(2*1 + 16) = 249.5 g/mol

4. moles of iron = 6.345/55.8 = 0.114 moles

mass of 0.45 mol of C2H6O = 0.45 * molar mass of C2H6O = 0.45 * (12*2 + 6*1 + 16)

= 20.7 g

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