When 1.00 g of a strong acid is dissolved in 100 mL of water the pH of the solution is 0.908. Assuming no change in volume when the solution is formed. Calculate the molar mass of the unknown acid?

Question

When 1.00 g of a strong acid is dissolved in 100 mL of water the pH of the solution is 0.908. Assuming no change in volume when the solution is formed. Calculate the molar mass of the unknown acid? (A) 20 g textmo textl-1  (B) 36 g textmo textl-1  (C) 63 g textmo textl-1  (D) 81 g textmo textl-1

Tardigrade Admin · Accepted Answer

pH=0.908=- log [H+] [H+]=0.1235 M=[Acid] m mol of acid = 12.35 Molar mass =(1000/12.35)=80.97≈ 81 g mol-1

Q. When 1.00 g of a strong acid is dissolved in 100 mL of water the pH of the solution is 0.908. Assuming no change in volume when the solution is formed. Calculate the molar mass of the unknown acid?

20 g $mo l^{- 1}$

36 g $mo l^{- 1}$

63 g $mo l^{- 1}$

81 g $mo l^{- 1}$

$p H = 0.908 = - lo g [H^{+}]$ $[H^{+}] = 0.1235 M = [A c i d]$ $m$ mol of acid = 12.35 Molar mass $= \frac{1000}{12.35} = 80.97 \approx 81 g m o l^{- 1}$

Q. When 1.00 g of a strong acid is dissolved in 100 mL of water the pH of the solution is 0.908. Assuming no change in volume when the solution is formed. Calculate the molar mass of the unknown acid?

20 g mol−1

36 g mol−1

63 g mol−1

81 g mol−1

pH=0.908=−log[H+] [H+]=0.1235M=[Acid] m mol of acid = 12.35 Molar mass =12.351000​=80.97≈81gmol−1

20 g $mo l^{- 1}$

36 g $mo l^{- 1}$

63 g $mo l^{- 1}$

81 g $mo l^{- 1}$

$p H = 0.908 = - lo g [H^{+}]$ $[H^{+}] = 0.1235 M = [A c i d]$ $m$ mol of acid = 12.35 Molar mass $= \frac{1000}{12.35} = 80.97 \approx 81 g m o l^{- 1}$