Calculate the molecular weight of a small protein if a 0.21-g sample dissolved i

Question

Calculate the molecular weight of a small protein if a 0.21-g sample dissolved in 191 mL of water has an osmotic pressure of 8.9 mmHg at 26 degree C. (R = 0.0821 L. atm (K middot mol)) A) 3.3 times 10^2 g/mol B) 3.0 g/mol C) 2.3 times 10^3 g/mol D) 3.0 times 10 5 g/mol E) 2.0 times 10^2 g/mol For the reaction IO^- (aq) + 5I^- (aq) + 6H^+(aq) rightarrow 3I_2(aq) + 3H_2O(l) the rate of disappearance of IO_3^- (aq) at a particular time and concentration is 2.5 times 10^-3 mol/(Ls). What is the rate of appearance of I_2(aq)? A) 2.5 times 10^-3 mol/(L middot s) B) 7.5 times 10^-2 mol/(L middot s) C) 8.3 times 10^-3 mol/(L middot s) D) 1.2 times 10^-3 mol/(L middot s) E) -7.5 times 10^-3 mol/(L middot s) What type of colloid is formed when a gas is dispersed in a solid? A) sol B) gel C) aerosol D) emulsion E) foam For a reaction of the form A rightarrow products, a plot of 1/[A] versus time is linear for a reaction of which order? A) second B) fourth C) third D) first E) zero

Explanation / Answer

1)

the formula for osmotic pressure () is:

= MRTi, where M is molarity, R is the gas constant (formmHg in this case), T is the temperature, and "i" stands for thenumber of dissociated particles there are per unit of the dissolvedsubstance - it's just 1 in this case, so it can be ignored.

So...

= MRT

8.9 = M(62.4)(299)

M = 4.77*10-4mol/L

Next, find the number of moles using the given volume and themolarity:

191mL * (4.77*10-4mol/1000mL) =9.11*10-5mol

For molecular weight, just divide the given mass by thenumber of moles:

0.21g/(9.11*10-5mol) = 2305.16g/mol,the final answer=2.3*10^3 g/mol

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