3. Consider the transport of galactose across a semi-permeable membrane from the extracellular space ([galactose] = 2 mM) to the cytosol ([galactose] = 10 mM) under biochemically standard conditions. (R = 0.08206 L atm/mol K = 8.314 J/mol  K) (15 pts.) If ATP hydrolysis liberates 30.5 kJ/mol at standard conditions, approximately how many grams of ATP (507.2 g/mol) would be required to transport 260 g of galactose from the extracellular space to the cytosol?

15.63.

Explanation:

So, from the question, we are given the following parameters or data or information for solving this particular Question and they are;

(1). "semi-permeable membrane from the extracellular space ([galactose] = 2 mM) to the cytosol ([galactose] = 10 mM) under biochemically standard conditions. (R = 0.08206 L atm/mol K = 8.314 J/mol  K) (15 pts.) "

(2). "If ATP hydrolysis liberates 30.5 kJ/mol at standard conditions".

(3). " (507.2 g/mol) would be required to transport 260 g of galactose from the extracellular space to the cytosol"

Step one: determine the difference to transport of galactose from extracellular space to cytosol.

This can be done by using the formula below:

Ratio = intracellular/ extracellular.

=> 10/2 = 5 mM.

Step two: so, from statement (2) and (3), the if 507.2 g/mol ATP = 30.5 kJ/mol, then 260 g/mol = ???

Therefore, 260 × 30.5 / 507.2 = 15.63 is the energy required.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm