1. If you have 10.0 mL of 3.0 M HCl and you want 0.50 M HCI. How much water in m

Question

1. If you have 10.0 mL of 3.0 M HCl and you want 0.50 M HCI. How much water in mL do you need to make the 0.50 M HCl solution? a, 50 mL b, 60 mL c. 70 mL d.. 80 mL Calculate the amount of mole) of acid titrated and what is the weight of malic acid? Balanced Chemical eq is: CeHoosa) 2 NaoHoa) NazC4H40s(ag) 2 H20eid) a. 0.0203 mol and 1.20% b, 0.122 mol and 2.01% c. 0.102 mol and 3.01% d. 0.0203 mol and 1.56% 3. What is the oxidation number of C in a compound, C2Hs d. +6 c. +5 4. You have 20.00 mL of 0.1005 M HCI in a flask and add an phenolphthalein. You then add a KOH solution of unknown molarity to a buret and records a initial reading as 1.50 mL. You add KoH solution to the HCi solution until a pale pink color is obtained and records a final buret reading as 22.00 mL. What is the molarity (M) of the KOH solution? a. 8.095 x 103 b, 9.805 x 103 c. 11.805 x 10 d. 10.905 x 10

Explanation / Answer

1-M1V1=M2V2

3x10 = 0.5 xV2

V2 = 3x10 / 0.5 = 3x 20 =60 ml

2-MW= 12x4+1x6+ 16x5= 48+6+ 80 = 134

mole of malic acid = 86.8/134 = 0.6477

mole of NaOH = 0.663x30.65/1000= 0.0203

2mole NaOH nutralise 1 mole malic acid

0.0203 mol   NaOH     = 0.0203/2= 0.01015=0.0102molemalic acid

wt of malic acid =134x0.0102=1.3668 %wt=11.02%

3- C2H8

Cx2 + 8xH = 0

2C+ 8x1=0

C=-4

4-V1(HCl)= 20ml M1(HCl)=0.1005M

V2( KOH)= 22-1.5=20.5

V1M1= V2M2

M2= V1M1/V2 = 20 x 0.1005 /20.5=0.9755x0.1005=0.0980487M KOH

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