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Elza [17]
3 years ago
8

What is the molarity of a solution that contains 25 g of HCl in 150 mL of solution??

Chemistry
2 answers:
oksian1 [2.3K]3 years ago
6 0

Answer : The Answer Is 4.6

Stels [109]3 years ago
3 0
To find molarity 
1) number of mol of solute.
Solute is HCl.
M(HCl)= 1.0+35.5 =36.5 g/mol

25g *1 mol/36.5 g = 25/36.5 mol HCl

2) Molarity is number of mole of the solute in 1 L solution.
150 mL = 0.150 L

(25/36.5 mol HCl )/(0.150 L) = 25/(36.5*0.150) ≈ 4.57≈4.6 mol/L


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3 0
2 years ago
A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

6 0
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