Answer to Question #317861 in Physical Chemistry for perfect

Question

Assume that mitochondria contain 0.47 Molar KCL and 0.015 Molar NaCl. Calculate the grams per liter, of a carbohydrate similar to glucose with a molecular weight 280 amu, that will have the same osmolarity as the mitochondria..

Answer 1

C(KCl) = 0.47 M;

C(NaCl) = 0.015 M;

O = i * C;

KCl = K⁺ + Cl^-;

NaCl = Na⁺ + Cl^-;

i(KCl) = i(NaCl) = 2;

O(KCl) = 2 * 0.47 = 0.94 osmol/L;

O(NaCl) = 2 * 0.015 = 0.03 osmol/L;

O(mitochondria) = O(KCl) + O(NaCl) = 0.94 + 0.03 = 0.97 osmol/L;

The simple formula of carbohydrate = C_x(H₂O)_y;

M(C_x(H₂O)_y) = 280 g/mol;

i(C_x(H₂O)_y) = 1;

Therefore,

C(C_x(H₂O)_y) = O(C_x(H₂O)_y)/i(C_x(H₂O)_y) = 0.97 mol/L;

C(C_x(H₂O)_y), g/L = C(C_x(H₂O)_y), mol/L * M(C_x(H₂O)_y) = 0.97 * 280 = 271.6 g/L.

Answer: 271.6 g/L.