Chemistry, 21.09.2019 01:30 LanaParrilla

12 instructions: • calculate both the volume and weight of each component to make the listed solutions. make sure to show your calculations for full credit. name both the volume or weight, units, and name of chemical for both the solute and the solvent. 1. 50ml of 1% acetic acid 2. 25ml of 10% sodium chloride 3. 10ml of 0.1% quinine sulfate 4. 1l of 5% sucrose 5. 60ml of 0.2% o-dianasidine in methanol 6. 50ml of 3% sodium hydroxide 7. 75ml of 1% sodium nitroferrocyanide

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Answer from: isbella29

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1). 50 ml of 1% acetic acid

Solute: acetic acid (Formula: CH₃COOH)

Solvent: water (H₂0), if nothing else is indicated, the solvent is always water)

1% acetic acid means that there are 1 gram of acetic acid in 100 ml of solution: 1 g CH₃COOH/100 ml solution. We need 50 ml of solution, thus we can divide into 2 both numerator and denominator in the relation as follows:

$\frac{1g CH3COOH}{100ml solution} = \frac{0.5 g CH3COOH}{50 ml solution}$

To prepare the solution: weigh 0.5 g of acetic acid and dissolve in 50 ml water.

2) 25 ml of 10% sodium chloride

Solute: sodium chloride (NaCl)

Solvent: water (H₂O)

The same as in 1), we divide in this case into 4 both numerator and denominator (to obtain the volume we need: 100ml/4= 25 ml)

$\frac{10 g NaCl}{100 ml solution} = \frac{2.5 g NaCl}{25 ml solution}$

To prepare the solution: weigh 2.5 g of NaCl and dissolve in 25 ml water.

3) 10 ml of 0.1% quinine sulfate

Solute: quinine sulfate (QS) (Formula: C₂₀H₂₄N₂O₂)

Solvent: water (H₂O)

$\frac{0.1 g QS}{100 ml solution} = \frac{0.01 g QS}{10 ml solution}$

To prepare the solution: weigh 0.01 g of quinine sulfate and dissolve in 10 ml water.

4) 1 L of 5% sucrose

Solute: sucrose (Formula: C₁₂H₂₂O₁₁)

Solvent: water (H₂O)

$\frac{5 g sucrose}{100 ml solution} = \frac{50 g sucrose}{1000 ml solution}$

To prepare the solution: weigh 50 g of sucrose and dissolve in 1000 ml water (1000 ml= 1 L).

5) 60 ml of 0.2% o-dianasidine in methanol

Solute: o-dianasidine (ODS) (Formula: C₁₄H₁₆N₂O₂)

Solvent: methanol (CH₃OH). Note that in this case it is indicated that the solvent is not water, but it is methanol.

$\frac{0.2 g ODS}{100 ml solution} =\frac{0.12 g ODS}{60 ml solution}$

To prepare the solution: weigh 0.12 g of ODS and dissolve in 60 ml methanol.

6) 50 ml of 3% sodium hydroxide

Solute: sodium hydroxide (NaOH)

Solvent: water (H₂O)

$\frac{3 g NaOH}{100 ml solution} = \frac{1.5 g NaOH}{50 ml solution}$

To prepare the solution: weigh 1.5 g of NaOH and dissolve in 50 ml water.

7) 75 ml of 1% sodium nitroferrocyanide

Solute: sodium nitroferrocyanide (NFC) (Formula: C₅FeN₆Na₂O)

Solvent: water (H₂O)

$\frac{1 g NFC}{100 ml solution} =\frac{0.75 g NFC}{75 ml solution}$

To prepare the solution: weigh 0.75 g of NFC and dissolve in 75 ml water.

Answer from: Quest

The density would be 1.622

Answer from: Quest

a electronegativity decreases beacuse the size of nucleus is bigger.

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