6 g of non volatile non electrolytic solute x is present in 1 litre and this solution is isotonic with 0.2925 (w/v) % Nacl solution(Assume Nacl completely dissociate and temperature of both solution is 27 degrees celcius) . molar mass of x is ?
Answers
Given:
The mass of x, w = 6 gm
The volume of x, V = 1 L
w/V % of NaCl = 0.2925 %
T = 27 °C
NaCl dissociates completely.
To Find:
The molar mass of x.
Calculation:
- We know that: NaCl ⇄ Na⁺ + Cl⁻
- As NaCl dissociates completely, i = 2/1 = 2
- 0.2925 % w/V ≡ 0.2925 gm NaCl in 100 ml water
- The molarity of NaCl, M(NaCl) = (0.2925 × 1000)/(58.5 × 100) = 0.05 M
- The molarity of x, M(x) = (6 × 1000)/(M.wt × 1000) = 6/M.wt
- Since solutions of x and NaCl are isotonic, we have:
Π(x) = Π(NaCl)
⇒ M(x) × R × T = i × M(NaCl) × R × T
⇒ 6/M.wt = 2 × 0.05
⇒ M.wt = 6/0.1
⇒ M.wt = 60 gm/mol
- So, the molar mass of x is 60 gm/mol
Given according to the question :
The mass of x, w = 6 gm
The volume of x, V = 1 L
w/V % of NaCl = 0.2925 %
T = 27 °C
NaCl dissociates completely.
To Find:
The molar mass of x.
Calculation:
- We know that: NaCl ⇄ Na⁺ + Cl⁻
- As NaCl dissociates completely, i = 2/1 = 2
- 0.2925 % w/V ≡ 0.2925 gm NaCl in 100 ml water
- The molarity of NaCl, M(NaCl) = (0.2925 × 1000)/(58.5 × 100) = 0.05 M
- The molarity of x, M(x) = (6 × 1000)/(M.wt × 1000) = 6/M.wt
- Since solutions of x and NaCl are isotonic, we have:
Π(x) = Π(NaCl)
⇒ M(x) × R × T = i × M(NaCl) × R × T
⇒ 6/M.wt = 2 × 0.05
⇒ M.wt = 6/0.1
⇒ M.wt = 60 gm/mol
- So, the molar mass of x is 60 gm/mol