Question

1- Calculate the following: A- Number of moles in 10 gram of Cr(NO3). B- Number of Atoms in 10 gram of Cr(NO3). C- Number of molecules in 10 gram of Cr(NO3). D- Number of grams in 10 atoms of Cr(NO3). 2. Find the % composition of all individuals in, KBrO. %K= %Br= %0=

Answer 1

Moleculae mass of GolN03)3 - 3 52+ 3462) 3 238g/molen No of moles = mass Molecular mass log = 0.042 mole E - 238g 1 mol 1 mol has 6.02x10 23 Betone moleculee ii 0.042 molee has = 2.02X.10% 3x 01042 2:53 x 1022 atosan moleculea 1 molecule has total 13 atome. i. Total number of atome = 13x2.53x1022 = 3.3 x 1023 atome c) - Number of molecules = 2.53X1022 molecules Number of moles - number Number of of atome. atoms is 1 mol. Number of atome in 1 mol = 13x6.02 x 1023 i No.of molee 1. 3x10 moles 100 13x6.02 X1023 mass = weight = No. of moles x molecular 1.3810-15 x 238 3 3.110-13a.

In KBr 07. KBYO. .Molos mass of KBxb2 = 39+80+ zlich 32319/mol. yok = No.of_atoms of_kx_atosuc mees of i Molecular mase of K Broz 16.900 1 x 39 XIUD = 231 7. Br = No of atoms of Br x_atomic maze of Br x 103 - Molecular mass of K Broz - 1x 80 231 x 100 = 3406 40 X 100 100 = No. of atome ofo x atomic mers of - Molecular mase of K Brot x 100 = 48.5.76 7 x 16 231 3. No. of oxygen atom per mole of Crim03)3 - = 9. No.of molee of ColN03)3_in_Log = 0.0 Atomic mass of each oxygen atom = 16g/mol ..i'. Mass of oxygen 9x 0,042 x 16 26.059:

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