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2. Determine the oxidation number for the indicated element in each of the following substances. a. So, b. Mno, c. HNO, d. Cr,0,2 4. (a ) Calculate the molarity of a solution made by dissolving 0.75g of Na, so, in water to form exactly 850 mL of solution. (b ) Calculate the volume of 1.50 M Na PO, in liters that contains 2.5 g of solute.

Answer 1

For determining oxidation number of indicated elements , we will consider (-2) charge of oxygen atom and (+1) charge of hydrogen atom.

We will take 'x' for the indicated atom.

we know that charge of a neural molecule is zero.

a. 502 30 +.2001 - 0 + 2(-2) = 0 OC - 4-0 a = +4 b. mnou x + 4 (0) = -1 20 + 4 (-2)= -1 2C-8 :-1 OC :-) + 8 x = +7 C. HNO, d. cozo, 2 (H) + OC + 200): 0 101) +20 +2 (-2)=0 x +1 -4:0 200 + 110) = -2 2x + 7(-2) = -2 2x -14-2 2x: -2+14 206:12 x = +6 x - 3 :0 2 = +3

So Oxidation number for the indicated elements are:

a. S = +4

b.Mn = +7

c.N = +3

d.Cr = +6

Now,

4(a)

4.ca? we have, w= 0.75 gram v=850 ml molecular weight of Nu,soy m.w: 2 (na) + I (S) + 410) = 2(23)+1 (32) + 4(16) - 46 + 32 + 64 = 142 go/mole Molurity cm) = 1000 * weight of solute m.w of solutex rolume in me = 1000 X 0.75 gm 142 gmx 850 ml = 6.21 *103 m

Molarity of a solution is 6.21× 10^-3 M.

(b)

Cb) we have W: 2.5 gm (M): 1.5 m V: 9. molecular weight of NY,DOY m.w: 3 (NU) + ICP) + 410) = 3 (23) +1 (31) + 4(16) = 69 +31 + 64 - 164 gol mole Now, molusinyrm): 1000 x weight of solute (w) m.w of solute volume (nl) 1, s n = ºs 1 0 0 0 1 2 4 3 1 x rolureime) 164 gr/mole . Yo Jq me (n) : lo 0 0 * 2 , S 164 X los = 10.16 m = 0.0102 Liten.

Volume is 0.0102 liter.

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