Question

3. The volume of blood plasma in an adult is 5.0 L, about 95% of which is water, and it has a density of 1.025 g/mL. The pressure at an underwater depth of 30.0 m is 4.1 atm (4.154325 bar). If a SCUBA diver descends to this depth, how many mg of N, (KH=6.0 x 10 mol/kg*bar) are dissolved in her blood? What percentage of N, comes out of the solution when the diver surfaces (at sea level)? Assume constant temperature. (6 pts.) ask kineret to explain

Answer 1

3.

Volume of water in the blood plasma = 5.0 (L) $\times$ $\frac{95}{100}$ = 4.75 L = 4.75(L) $\times$ = 4750 mL

1000ml 1L

density = 1.025 g/mL

then mass of water = volume $\times$ density

= ( 4750 mL $\times$ )

1.0259 mL

= 4869 g

= 4869 (g) $\times$   

1Kg 1000g

= 4.869 Kg

now

Henry's law equation is

Cg = KH $\times$ Pg

given , Pg at 30 m depth = 4.154325 bar ; KH  = 6.0$\times$10-4 mol/kg bar

Then , concentration of N2 in mol/Kg is

Cg =  6.0$\times$10-4 (mol/kg bar) $\times$ 4.154325 (bar)

= 0.002492 mol/Kg

now, in 4.869 Kg of water moles of N2 present

= 4.869 (Kg) $\times$ 0.002492 mol/Kg

= 0.01213

mass of N2 = moles$\times$ molar mass = 0.01213 (mol)  $\times$ 28 (g/mol) = 0.340 g

= 0.340 (g) $\times$

1000(mg) 19)

= 340 mg

therefore mass of N2 dissolved in her blood at 30 m depth =  340 mg

again at sea level pressure ( Pg) = 1.01325 bar

now, Cg =  6.0$\times$10-4 (mol/kg bar) $\times$ 1.01325 (bar)

= 0.000608 mol/Kg

now, in 4.869 Kg of water moles of N2 present

= 4.869 (Kg) $\times$ 0.000608 mol/Kg

= 0.00296

mass of N2 = moles$\times$ molar mass = 0.00296 (mol)  $\times$ 28 (g/mol) = 0.0830g

= 0.0830 (g) $\times$

1000(mg) 19)

= 83  mg

now, total amount of N2 comes out

= (340 - 83.0) mg

= 257mg

then , percentage of N2 comes out from the solution when the driver surfaces ( at sea level)

=

massof N2 comes out x 100 mass of 12 dissolved at 30m depth

=

257(mg) x 100 340(mg)

= 75.6 %