Dear Student, â€‹1 molecule of Na2SO4 7H2O contains 11 atoms of Oand, 1 mol of Na2SO4 7H2O contains 6 023×1023 molecules of Na2SO4 7H2O So, 1 mol of Na2SO4 7H2O contains 6 023×1023×11 atoms of Oor, 6 023×1023 atoms of O is present in 16 023×1023×11×6 023×1023= 0 09 mol of Na2SO4 7H2ONow, moles=massmolar massor, mass = moles×molar mass = 0 09×268 = 24 12 g

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solve Q 5 Q.5 Q.6 Q.8 dtomstnamu t 00 tßiifMassof100atoms of ingrn. ; '4-vfio. ofmolecules & atoms in 54 gm H20. 0<0. of atoms in 88 gm C02. What will bethemæssofone atom ing? X!' 65 Y/O Calculate massofO atoms in 6 gmCH3COOH? Calculate mass of water present in 499 gm CuS04.5H20 ? (Atomic mass — Cu = 63.5, S = 32, O = 16, H = 1) What mass ofNa2S04.7H20 contains exactly 6.023 x 1022 atoms ofoxygen? SIOUCUÅOMETRY How many gmofHCl is needed for complete ractionwith69.6gmMn02?. HCI + Mn02—» MnC'12 + H20 + Nitric acid is manufactured by the Ostwald process, in which nitrogen dioxide 3 N02 (g) + H20 (i) 2 HN03 (aq) + NO (g) How many grams of nitrogen dioxide are required in this reaction to produce 25.; Flourine reacts with uranium to produce uranium hexafluoride, IJF6, as

Dear Student,

1 molecule of {Na}_{2} {SO}_{4} . 7 H_{2} O contains 11 atoms of O and, 1 mol of {Na}_{2} {SO}_{4} . 7 H_{2} O contains 6.023 \times 10^{23} molecules of {Na}_{2} {SO}_{4} . 7 H_{2} O . So, 1 mol of {Na}_{2} {SO}_{4} . 7 H_{2} O contains 6.023 \times 10^{23} \times 11 atoms of O or, 6.023 \times 10^{23} atoms of O is present in \frac{1}{6.023 \times 10^{23} \times 11} \times 6.023 \times 10^{23} = 0.09 mol of {Na}_{2} {SO}_{4} . 7 H_{2} O Now, moles = \frac{mass}{molar mass} or, mass = moles \times molar mass = 0.09 \times 268 = 24.12 g

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