Question

1) Calculate the number of molecules in the following: a) 0.25 mol of glucose (C6H1206) b) 1.4 g of HCI c) 0.35 g of NH3 2) What is the mass of the following: a) 5.00x10" molecules of HI b) 0.0250 mol of Al(SO4)3 c) 1.00x10% atoms of Na 3) How many are there: a) moles in 2.50g of K PO. b) grams in 0.034 mole of HCI c) moles in 1.25g of NazS 4) Calculate a) How many g of Ca are in 1.55g of Ca:P b) How many g of Al,O, contain 7.50g of Al?

Answer 1

1. (a) 1.5055×10²³ molecules

(b) 2.31×10²² molecules

(c) 1.24×10²² molecules

2.(a) 10.6 g (b) 8.55 g (c) 3.82 g

3. (a) 0.0118 moles (b) 1.24 g (c) 0.0160 moles

4. (a) 1.02 g (b) 14.2 g

23 Ind = 6.022 x 103 molecules 1. a) So, number of molecules en 0.25 and glucose = 0.25 mol x 6.022x10 molecules/med 23 11.5055810 molecules (6) Nolas mass of HU = (1,0 + 35.5)g/mol = 36'5 g/mol No. of moles = 1.49 = 14 mon 36.5 9md 365 Number of molecules = 1.4 x 6.022x10 molecules 3615 23 с.231 х 10 - ) ео) 22 = /2.31 x 10 molecules) c) Molar mass of Mt₂ = (14.0+ 3x1000) g/mol = 170 g/mol Moles of NH3= 0.35 g = 0.35 mo 1700 g/mo no 0.35 23 Number of 17.0 molecules = 0.35 x6:022 x 10 insecules = 0.124 x10 molecules = /1.24x102 molecules C anned with Scanner

2. a) Moles of A a) Moles of Hi molecules = 5.00x 18 molecules 61022x10 23 molecules/mol = 0.083 mol Molar mass of HI = (1.00 + 126 99 g/mol = 127.9 g/mol Now mass = 0.083 mol x 127.9 g/mol = 10.6 g 1 b) Molar mass of Al2(S2), = [ax 27.0+ 3(32+4x16) gimon = 342 g/mol PPPPPP Now mass = 0.0250m x 342 g/men = 8.5591 10 x 10 c) Number of woles of Na atoms=- 6'022X1023 = 0.166 mol A Molar mass of Na = 23:0 g/mon So, mass = 0.166 mol x 23:0 g/mes = 3.8189 - 13.829) Scanned with am Scanner

3.a) Molen mass og K3 PO4 = (3x 39.1+ 30197+4x1600)gmos = 212:27g/mol Number of roles = 2.509 510-0118 moles 212-27 g/mol 6) Molar mass of HU = 365 g/mol Mass g 0.034 mele = (0-034 x 365) = 11:249] c) Molar mass of Nars = (2x2300 +32) g/mol = 78.0 g/med Number of moles n = - 1.259 78.0 ghmon -0.0150 moler L =10:0160 endles 4 (2) Molecular Mass of Cazpe in grams = (3x40-0 + 2x30.97) = 181.949 Mass of 3 Ca atoms = 3x40-09 - 1209 This means 181.94 g Cazle contains 120g Ca. So, mass of ca in 155g Cash = 120gea 181.949* 1.559 = 11:02 g/ca hned with Scanner

6) Molecular mass of Al2O3 in grams - (2x 27.0 + 3x16•0) g = 1029 Mass of 2 Alatoms = 2x 27-09 354-09 This means, 1029 of Al2 O3 contains 5400g Al. So, mass of Al2O3 that contains 7.50 g Al - 1029 A1203 2x7.50g Al A 5400g AT 114.2g Al2O3 Scanned with CamScanner