Question

Hy 1.008 664,427 0 15.964 18. A biochemical engineer isolates a bacterial gene fragment and dissolves a 20.6. mg sample of the material in enough water to make 41.5 mL of solution. The osmotic pressure of the solution is 0.340 torr at 25.0°C. (a) What is the molar mass of the gene fragment (5 pts)? 2mci we6 .0206 s .0415 C= .0206 ms.

Answer 1

Here is the Solution:

Given data : mass of the bacterial gene = 20.0 mg (m) 20.6 mg x- 0.0206 g 1000 mg volume = 41.5 mx x 12 = 0.0415L (²) 1000 ml Osmotic pressure of the solution = 0.340 torr . 1 atm 0.340 torr x atm = 4.474× 10 -4 atm 760 torr * Molarity of the solution , m= I T = osmotic pressure 4.474x10 -4 atat R = Gas constant 0.0821 Liatm/malik (0.0821 Lato/mol•K)x (298.15%) T = temperature M = 1.83 x 10-5 mol/ 25+273.15 = 298.15K M = N0 0 ) Molar mass = mass of bacterial gene molarity volume 0.0206 Molar mass = g x 1.83x10-5 mol l 0.0415K Molar mass = 2.71 Hence, fragment is: the molar mass of the gene 2171 X 104 g/mol

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