Answer:
A. 7,20x10⁻³mol NaF
B. 4,33x10²¹ F⁻ ions
C. 1,88x10⁻³g Na⁺
D. 3,1x10²² formula units
E. 132,08g/mol
F. 15,9g of CHCl₃; 0,293 mol of CHCl₃
Explanation:
A. 119g × (0,254% / 100) = 0,302g of NaF × (1mol / 41,99g) = <em>7,20x10⁻³mol NaF</em>
B. 7,20x10⁻³mol NaF ≡ 7,20x10⁻³mol F⁻ × (6,022x10²³ ions / mol) = <em>4,33x10²¹ F⁻ ions</em>
C. 1,35g × (0,254% / 100) = 0,00343g of NaF × (22,99g Na⁺ / 41,99g NaF) = <em>1,88x10⁻³g Na⁺</em>
D. 119g × (4,4% / 100) = 5,236g of KNO₃ × (1mol / 101,1g) = 0,0518mol × (6,022x10²³ F.U / 1mol) = <em>3,1x10²² formula units</em>
E. Formula of ammonium sulfate is (NH₄)₂SO₄. Molar mass is obtained thus:
N: 2N×(14g/mol) = 28g/mol
H: 8H×(1,01g/mol) = 8,08g/mol
S: 1S×(32g/mol) = 32g/mol
O: 4O×(16g/mol) = 64g/mol
Molar mass: 28 + 8,08 + 32 + 64 = <em>132,08g/mol</em>
F. Molar mass of chloroform is 119,38g/mol, thus:
0,133 mol × (119,38g/mol) = <em>15,9g of CHCl₃</em>
35,0g × (1mol / 119,38g) = <em>0,293mol of CHCl₃</em>
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I hope it helps!
