Question

How much of water should be added to 24ml of alcohol to obtain a 12% alcohol solution?

Q2. How much mass of sodium sulphate is required to prepare 20% solution in 200g of water?

Q3. solubility of ammonium chloride is 37g at 293 K.What mass of ammonium chloride is required to make a saturated solution in 15g of water at 293K​

Answer 1

For 1st question

Let the required volume of water (solvent) we 'x' mL .

Total volume of solution :-

= (24 + x) mL

Now, for v/v % concentration, we have :-

⇒ Vol of solute/Vol of solution × 100

⇒ 24/(24 + x) × 100 = 12

⇒ 2400 = 12(24 + x)

⇒ 2400 - 288 = 12x

⇒ x = 2112/12

⇒ x = 176 mL

For 2nd question

Let the required mass of Sodium Sulphate (solute) be 'x' grams .

Total mass of the solution :-

= (x + 200) g

For w/w % concentration, we have :-

⇒ Mass of solute/Mass of solution × 100

⇒ x/(x + 200) × 100 = 20

⇒ 100x = 20(x + 200)

⇒ 100x - 20x = 4000

⇒ x = 4000/80

⇒ x = 50 grams

For 3rd question

Solubility of Ammonium Chloride is 37g at 293 K . This means at 293 K, 100 g of water can dissolve 37g of ammonium chloride.

Now, we can calculate our answer using unitary method.

Required mass of Ammonium Chloride :-

= (15 × 37)/100

= 555/100

= 5.55 grams

Answer 2

Answer:

Question :-

1》

• How much water should be added to 24mL of alcohol to obtain a 12% of alcohol solution.

Given :-

• In the question given that the 24mL of alcohol to obtain a 12% of alcohol solution.

To find :-

• Volume of water in the given solution.

Answer :-

• Here we get answer as volume of water in the given alcohol solution is 170mL.

Solution :-

• Lets take the volume of water in the given alcohol solution as x mL.
• Here the total volume becomes as,
• =(24+x)mL.
• To get the volume of water in the given alcohol solution by

• =Volume of solute / volume of solution×100
• Then we get the answer.
• Now applying all the values we get,

• $\frac{24}{24 + x} \times 100 = 12$
• 2400=12 (24+x)
• 2400=288+12x
• 2400-288=12x
• x=2112/12
• x= 176mL.

Therefore ,

• The volume of water given alcohol solution is 176mL.

2》

• How much mass of sodium sulphate is required to prepare 20% of solution in 200g of water.

Given :-

• Here given that mass of sodium sulphate is required to prepare 20% of solution in 200g of water.

To find ;-

• The required mass.

Answer :-

• The required mass is 50 grams .

Solution :-

• Consider,

• The required mass of sodium sulphate be x grams .
• Total mass will be
• x+200g.
• Here to get the required mass of sodium sulphate is
• = mass of solute/ mass of the given solution ×100.
• Now applying all the values we get,

• x/x+200×100=20

• 100x=20 (x+200)
• 100x=20x+4000
• 100x-20x=4000
• 80x=4000
• x=4000/80
• x=50grams.

Therefore ,

• The required mass of sodium sulphate is 50 grams .

3》

• Solubility if ammonium chloride is 37g at 293k.what mass of ammonium chloride is required to a saturated solution in 53g of water at 293k.

Answer :-

• The required mass of ammonium chloride is 5.55 grams .

Solution :-

• Required mass of ammonium chloride = 15×37/100
• =15×37/100
• =555/100
• =5.55 grams.

Used formulae :-

• Unitary method.

Therefore ,

• The required mass of ammonium chloride is 5.55 grams.

Hope it helps you.

Thank you .