Question

Solution(s) containing 30 g $C{H}_{3}COOH$ is/are:

• A. 50 g of 70 % (w/v) $C{H}_{3}COOH$
  [$d_{sol}=1.4g/mL]$
• B. 50 g of 10 M $C{H}_{3}COOH$
  [$d_{sol}=1g/mL]$
• C. 50 g of 60 % (w/w) $C{H}_{3}COOH$
• D. 50 g of 10 m $C{H}_{3}COOH$
  

Answer 1

Answer: (B), (D)

The correct options are
B 50 g of 10 M $C{H}_{3}COOH$
[$d_{sol}=1g/mL]$
D 50 g of 10 m $C{H}_{3}COOH$


70 % (w/v) means 70 g of solute is present in 100 mL of solution.
Since, density of solution is given,
volume of 50 g of $C{H}_{3}COOH$ = $\frac{50g}{1.4g/mL}$ = $35.71mL$
70 % (w/v) will contain $35.71\times \frac{70}{100}$ = 25 g

10 M solution means 10 moles of solute in 1 L of the solution.
Given 50 g of the solution with density 1 g/mL, so volume of the solution is 50 mL = 0.05 L.
We know,
Molarity = $\frac{\text{number of moles of solute}}{\text{volume of the solution in L}}$
So, 10 M = $\frac{\text{number of moles of solute}}{\text{0.05}}$
number of moles of solute = $10\times 0.05$ = 0.5 mol
amount of $C{H}_{3}COOH$ present = $0.5\times 60$ = 30 g

60% (w/w) means 60 g of solute is present in 100 g of solution
So, 50 g of solution will contain $\frac{60}{100}\times 50$ = 30 g

10 m solution means 10 moles of solute is present in 1 kg of the solution.
Molality = $\frac{\text{number of moles of solute}}{\text{mass of the solution in kg}}$
number of moles of solute = $10m\times 0.05kg$ = 0.5 mol
amount of $C{H}_{3}COOH$ = $0.5\times 60$ = 30 g